Codeforces Round #451 (Div. 2) A-B

A：

A. Rounding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, Vasya considers it already rounded.

For example, if n = 4722 answer is 4720. If n = 5 Vasya can round it to 0 or to 10. Both ways are correct.

For given n find out to which integer will Vasya round it.

Input

The first line contains single integer n (0 ≤ n ≤ 109) — number that Vasya has.

Output

Print result of rounding n. Pay attention that in some cases answer isn't unique. In that case print any correct answer.

Examples

input

5

output

0

input

113

output

110

input

1000000000

output

1000000000

input

5432359

output

5432360

Note

In the first example n = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.

int main()
{
    int n;
    cin>>n;
    if(n==0)
		cout<<"0"<<endl; 
	else
	{
	    vector<int>Q;
	    Q.clear();
	    int x=n;
	    int sum=0;
	    while(x)
	    {
	        Q.push_back(x%10);
	        x/=10;
	    }
	    int len=Q.size();
	    if(Q[len-1]==0)
	        cout<<x<<endl;
	    else
	    {
	        int z=10-Q[0];
	        if(z>Q[0])
	        {
	            printf("%d\n",n-Q[0]);
	        }
	        else
	            printf("%d\n",n+z);
	
	    }		
	}

    return 0;
}

123

B:

B. Proper Nutrition

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.

Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.

In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.

Input

First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.

Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.

Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.

Output

If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).

Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.

Any of numbers x and y can be equal 0.

Examples

input

7
2
3

output

YES
2 1

input

100
25
10

output

YES
0 10

input

15
4
8

output

NO

input

9960594
2551
2557

output

YES
1951 1949

Note

In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.

In second example Vasya can spend exactly n burles multiple ways:

• buy two bottles of Ber-Cola and five Bars bars;
• buy four bottles of Ber-Cola and don't buy Bars bars;
• don't buy Ber-Cola and buy 10 Bars bars.

In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.

扩展欧几里得： 求ax+by=z;

注意要非负数解

#include<iostream>  
#include<cstring>  
#include<cstdio>  
#include<malloc.h>  
using namespace std;  
typedef long long ll;  
  
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
  
int main()  
{  
    ll a,b,x,y,c;  
    cin>>c>>a>>b;  
    ll eg=exgcd(a,b,x,y);  
    if(c%eg)  
        cout<<"NO"<<endl;  
    else  
    {  
        x=x*c/eg;  
        x=(x%(b/eg)+b/eg)%(b/eg); 
        y=(c-x*a)/b;
        if(x<0|y<0)
        	cout<<"NO"<<endl;
        else
        {
          	cout<<"YES"<<endl;
      		cout<<x<<" "<<y<<endl;      	
		} 

    }  
    return 0;  
}