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.
ax+by=cax+by=c
set d=gcd(a,b)d=gcd(a,b), then it has a solution if and only if d|cd|c is
solved as follows:
Use extended Euclidean to find the solution of ax0+by0=dax0+by0=d
Then a(x0∗c/d)+b(y0∗c/d)=ca(x0∗c/d)+b(y0∗c/d)=c,
so the special solution is x′=x0∗c/d ,y′=y0∗c/dx′=x0∗c/d,y′=y0∗c/d
and general solution = special solution + homogeneous solution
The homogeneous solution is the solution of the equation ax+by=0 ax+by=0
Therefore, the general solution is x=x′+k∗b/d,y=y′−k∗a/dk∈z
For x, first multiply that multiple, and then x obtains the remainder as b/d, and x takes the remainder to obtain an x, and then in the case of this remainder, a y can be obtained, and it is judged whether the current y is positive
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <stack>
#include <map>
#include <unordered_map>
#include <vector>
#include <cmath>
#include <ext/rope>
#include <bits/stdc++.h>
using namespace std;
#define gt(x) x = read()
#define int long long
#define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
typedef pair<int, int> PII;
typedef unsigned long long ULL;
inline int read(int out = 0)
{
char c;
while((c=getchar()) < 48 || c > 57);
while(c >= 48 && c <= 57) out=out*10+c-48,c=getchar();
return out;
}
const int N = 1010;
const int M = 35;
const int mod = 1e9 + 7;
const int PP = 131;
const double eps = 1e-10;
int n;
int a, b;
int x, y;
int exgcd(int a, int b, int &x, int &y)
{
if (!b)
{
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
signed main(){
gt(n), gt(a), gt(b);
int d = exgcd(a,b, x, y);
// cout << d << endl;
if (n % d) cout << "NO" << endl;
else{
x *= (n / d);
int mod = (b / d);
x= (x % mod + mod) % mod;
y = (n - x * a) / b;
if (x < 0 || y < 0){
cout << "NO" << endl;
}
else{
cout << "YES" << endl;
cout << x << " " << y << endl;
}
}
return 0;
}