Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.

Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$

Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?

Input

The first line contains four integers — $A$

Output

If a distribution of $N$

If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$

Examples

Input

10 10 5 20

Output

Input

2 2 0 4

Output

-1

Input

2 2 2 1

Output

-1

Note

The first sample describes following situation: $5$

In the second sample $2$

The third sample describes a situation where $2$

$2$

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
int main(){
    int a,b,c,i,n,d;
    cin>>a>>b>>c>>n;
    d=n-(a+b-c);    
    if(c>a){
        cout<<"-1"<<endl;
    }
    else if(c>b){
        cout<<"-1"<<endl;
    }
    else if(d<1){
        cout<<"-1"<<endl;
    }
    else{
        cout<<d<<endl;
    }
    return

View Code

B - B

CodeForces - 991B

Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.

The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$

This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$

Help Vasya — calculate the minimum amount of lab works Vasya has to redo.

Input

The first line contains a single integer $n$

The second line contains $n$

Output

Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$

Examples

Input

3
4 4 4

Output

Input

4
5 4 5 5

Output

Input

4
5 3 3 5

Output

Note

In the first sample, it is enough to redo two lab works to make two $4$

In the second sample, Vasya's average is already $4.75$

In the second sample Vasya has to redo one lab work to get rid of one of the $3$

$3$

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
int main(){
    int n,i,a[1000],sum=0,k=0;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>a[i];
        sum+=a[i];
    }
    if(sum*1.0/n>=4.5){
        cout<<"0"<<endl;
    }
    else{
        sort(a,a+n);
        for(i=0;i<n;i++){
            sum=sum-a[i]+5;
            k++;
            if(sum*1.0/n>=4.5){
                cout<<k<<endl;
                break;
            }
            a[i]=5;    
        }
    }
    return 0;
}

View Code

C - C

CodeForces - 991C

After passing a test, Vasya got himself a box of $n$

This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$

If the amount of candies in the box is not divisible by $10$

Your task is to find out the minimal amount of $k$

Input

The first line contains a single integer $n$

Output

Output a single integer — the minimal amount of $k$

Example

Input

Output

Note

In the sample, the amount of candies, with $k = 3$

$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$

In total, Vasya would eat $39$

$39$

Vasya有n个糖果，在开始的时候 Vasya 选择了一个整数k，表示他每天会吃k个糖果，Petya想偷吃一部分糖果，他每天会吃当前数量的10%(下取整)的糖果

输出最小的k，使得Vasya至少吃掉一半的糖果。

注意：

若Vasya吃糖果时数量不满k，则Vasya会全吃掉。

若Petya吃糖果时数量不满10个，则Petya不会吃糖果。

用二分查找，二分需要满足两种结构：第一，解可能存在于一个固定的区间；第二，这个区间必须具有单调性。

这里的二分区间就是

#include<bits/stdc++.h>
using namespace std;

//typedef long long ll;
long long n;
long long check( long long x )
{
    long long now, sum;
    now = n, sum = 0;
    while( now )
    {
        if( now <= x )
        {
            sum += now;
            break;
        }//如果剩下的糖小于k时
        now -= x;
        sum += x;
        now -= now / 10;
    }
    return sum;//sum是指Vasya吃的糖的数量。
}
int main()
{

    while( cin >> n )
    {
        long long l = 1, r = n, mid;
        while( l < r )
        {
            mid = ( l + r ) / 2;
            if( check(mid) >= ( n + 1 ) / 2 )//取中间的mid的值为k是vasya吃的糖比Petya的多时
            {
                r = mid;
            }
            else
            {
                l = mid + 1;
            }
        }
        cout << l << endl;
    }
    return 0;
}

View Code

posted @ 2021-02-07 14:09 nanmoon 阅读(111) 评论(0) 编辑收藏举报

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