Educational Codeforces Round 21(A.暴力,B.前缀和,C.贪心)

A. Lucky Year

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.

You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.

Input

The first line contains integer number n (1 ≤ n ≤ 10⁹) — current year in Berland.

Output

Output amount of years from the current year to the next lucky one.

Examples

Input

4

Output

1

Input

201

Output

99

Input

4000

Output

1000

Note

In the first example next lucky year is 5. In the second one — 300. In the third — 5000.

题目链接：http://codeforces.com/contest/808/problem/A

分析：题目意思是要将下一个数字变为只有1个非0的数，打表即可，要注意开ll，否则会WA！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n;
    while(scanf("%lld",&n)!=EOF)
    {
        if(n>=0&&n<=9)
        cout<<1<<endl;
        else if(n>=10&&n<=99)
            cout<<(n/10+1)*10-n<<endl;
        else if(n>=100&&n<=999)
            cout<<(n/100+1)*100-n<<endl;
        else if(n>=1000&&n<=9999)
            cout<<(n/1000+1)*1000-n<<endl;
        else if(n>=10000&&n<=99999)
            cout<<(n/10000+1)*10000-n<<endl;
        else if(n>=100000&&n<=999999)
            cout<<(n/100000+1)*100000-n<<endl;
        else if(n>=1000000&&n<=9999999)
            cout<<(n/1000000+1)*1000000-n<<endl;
        else if(n>=10000000&&n<=99999999)
            cout<<(n/10000000+1)*10000000-n<<endl;
        else if(n>=100000000&&n<=999999999)
            cout<<(n/100000000+1)*100000000-n<<endl;
        else if(n>=1000000000&&n<=9999999999)
            cout<<(n/1000000000+1)*1000000000-n<<endl;
    }
    return 0;
}

B. Average Sleep Time

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days!

When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last n days. So now he has a sequence a₁, a₂, ..., a_n, where a_i is the sleep time on the i-th day.

The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider k consecutive days as a week. So there will be n - k + 1 weeks to take into consideration. For example, if k = 2, n = 3 and a = [3, 4, 7], then the result is .

You should write a program which will calculate average sleep times of Polycarp over all weeks.

Input

The first line contains two integer numbers n and k (1 ≤ k ≤ n ≤ 2·10⁵).

The second line contains n integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵).

Output

Output average sleeping time over all weeks.

The answer is considered to be correct if its absolute or relative error does not exceed 10^- 6. In particular, it is enough to output real number with at least 6 digits after the decimal point.

Examples

Input

3 2 
3 4 7

Output

9.0000000000

Input

1 1 
10

Output

10.0000000000

Input

8 2 
1 2 4 100000 123 456 789 1

Output

28964.2857142857

Note

In the third example there are n - k + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.

题目链接：http://codeforces.com/contest/808/problem/B

分析：计算上升和下降的序列数之和，再除以n-k+1即可，相当于求其中每k个数的和的平均值！

下面给出AC代码：

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll n,k;
double avg;
ll ans;
ll s[200005];
int main()
{
    cin>>n>>k;
    for(int i=1;i<=n;i++)
        scanf("%I64d",s+i);
    ll c=0;
    for(int i=1;i<=n;i++)
    {
        if(i<=n-k+1)
        {
            if(c<k)
                c++;
            ans+=c*s[i];
        }
        else
        {
            if(c>n-i+1)
                c--;
            ans+=c*s[i];
        }
    }
    avg=ans*1.0/(n-k+1);
    printf("%0.12lf",avg);
  return 0;
}

C. Tea Party

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumes a₁, a₂, ..., a_n. His teapot stores w milliliters of tea (w ≤ a₁ + a₂ + ... + a_n). Polycarp wants to pour tea in cups in such a way that:

• Every cup will contain tea for at least half of its volume
• Every cup will contain integer number of milliliters of tea
• All the tea from the teapot will be poured into cups
• All friends will be satisfied.

Friend with cup i won't be satisfied, if there exists such cup j that cup i contains less tea than cup j but a_i > a_j.

For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.

Input

The first line contains two integer numbers n and w (1 ≤ n ≤ 100, ).

The second line contains n numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100).

Output

Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.

If it's impossible to pour all the tea and satisfy all conditions then output -1.

Examples

Input

2 10 
8 7

Output

6 4 

Input

4 4 
1 1 1 1

Output

1 1 1 1 

Input

3 10 
9 8 10

Output

-1

Note

In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.

题目链接：http://codeforces.com/contest/808/problem/C

题意：n个茶杯，每个茶杯有容量。现在给一壶茶，总量为w。希望倒茶满足条件：每杯茶要超过容量的一半，并且w被倒光，茶杯内的茶水为整数，容量大的杯子内的茶不允许比容量小的杯子内的茶水少，特判不满足情况，然后将茶水给每一杯倒至一半以上。然后按照容量从大到小每次倒一个单位，直到倒完为止。

分析：贪心求解！
下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
class Tea
{
    public:
        int a,w,id;
};
Tea t[105];
int n,w;
int cmp1(Tea a,Tea b)
{
    return a.a>b.a;
}
int cmp2(Tea a,Tea b)
{
     return a.id<b.id;
}
int main()
{
    cin>>n>>w;
    for(int i=0;i<n;i++)
    {
        cin>>t[i].a;
        t[i].id=i;
    }
    sort(t,t+n,cmp1);  //杯子大到小排序
    for(int i=0;i<n;i++)
    {
        w-=(t[i].a+1)/2; //每个杯子至少装一半
        t[i].w=(t[i].a+1)/2;   //保存已经装了多少
    }
    for(int i=0;i<n;i++)
    {
        int x=t[i].a-t[i].w;
        if(w>=x)
        {
            t[i].w+=x;
            w-=x;
        }
        else if(w>0)
        {
            t[i].w+=w;
            w=0;
        }
        else
            break;
    }
    sort(t,t+n,cmp2);  //按照id 排序 排回原位
    if(w<0)
        cout<<-1;
    else
    for(int i=0;i<n;i++)
        cout<<t[i].w<<" ";
    return 0;
}