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.

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.

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.

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.

代码：

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 #include<cmath>

 5 #include<algorithm>

 6 using namespace std;

 7 typedef long long ll;

 8 int main(){

 9     ll n,a,b;

10     while(~scanf("%lld",&n)){

11         scanf("%lld%lld",&a,&b);

12         int flag=0;ll cnt,num;

13         for(int i=0;;i++){

14             num=n-i*a;

15             if(num<0)break;

16             if(num%b==0){flag=1;cnt=i;break;}

17         }

18         if(flag==1){

19             printf("YES\n");

20             cout<<cnt<<" "<<num/b<<endl;

21         }

22         else cout<<"NO"<<endl;

23     }

24     return 0;

25 }