CodeForce Proper Nutrition

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本文探讨了在有限的预算下，如何购买不同价格的商品，以达到预算恰好用完的目标。通过列举多个实例，如购买Ber-Cola和Barsbars，展示了如何通过算法找到可能的购买组合。文章还提供了AC代码示例，帮助读者理解实现过程。

Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs bburles. 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.

AC代码：

//Proper Nutrition 
#include<iostream>

using namespace std;

typedef long long ll;
 
int main() 
{
	ll n,a,b;
	while(~scanf("%lld",&n)){
		scanf("%lld %lld",&a,&b);
		int flag = 0;
		ll cnt,num;
		for(int i=0; ; i++){
			num = n - i * a;//枚举 
			if(num < 0){
				break;
			}
			if(num % b == 0){
				flag = 1;
				cnt = i;// x
				break;
			}
		}
		
		if(flag == 1){//满足条件 
			cout << "YES" << endl;
			cout << cnt << " " << num/b << endl;
		}else{
			cout << "NO" << endl;
		}
	}
	return 0;
}
/*

n = x*a + y*b

*/

值得参考的代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,a,b;
	while(~scanf("%d%d%d",&n,&a,&b))
	{
		bool flag = false;
		for(int i = 0; i * a <= n; i++)
		{
			if((n - a * i) % b == 0)
			{
				flag = true; printf("YES\n%d %d\n",i, (n - a * i) / b);
				break;
			}
		} 
		if(flag == false) printf("NO\n");
	}
	
	return 0;
}
/*
7
2 3
100
25 10
15
4 8
9960594
2551 2557
*/