[Codeforces 1244C] The Football Season（扩展欧几里得）

xw+yd=p,x+y+z=n求x,y,z

首先尽量让他赢p/w次，剩下m=p%w分，剩下的分给平，m/d个平，为什么不先分给平呢？w比较大，所以这样x+y比较小，不容易让z<0。

可是m%d不一定是0，这时候要调整，+-w，+-d能怎么调呢？就要用到扩展欧几里得算法了，aw+bd=gcd(w,d)，求出a,b,gcd.如果m%d%gcd不是0，那么无解，+-w，d调整的最小单位就是gcd。否则加上a*（m%d/gcd）个w和b*（m%d/gcd）个d，使得和为p。

还要注意，增加b/gcd个w同时减少w/gcd个d和是不变的，在胜利次数<0时可以这么调整。

#include 
#include 
#include 
#include 
 
using namespace std;

typedef long long LL;

void exgcd(LL a, LL b, LL& x, LL& y, LL& c) {
    if(!b) {y = 0; x = 1; c = a; return;}
    exgcd(b,a%b,y,x,c); y -= a/b*x;
}
  
int main() {
	LL n,p,w,d;
	cin >> n >> p >> w >> d;
	// wx+dy = gcd w,d
	LL x,y,c;
	exgcd(w,d,x,y,c);
	LL v1 = p/w,v2 = 0,m = p%w;
	v2 = m/d; m = m%d;
//cout << x << " " < t2) {
				v1 += t2;
			    v2 -= t1;
			    if (v2 < 0) {
				  puts("-1");
				  return 0;
			    }
			} else if (t1 < t2) {
				v1 -= t2;
			    v2 += t1;
			    if (v1 < 0) {
			 	  puts("-1");
				  return 0;
			    }
			} else {
				puts("-1");
				return 0;
			}
		}
		if (v1>=0 && v2>=0 && n-v1-v2>=0) {
			cout << v1 << " " << v2 << " " << n-v1-v2 << "\n";
			return 0;
		}
	}
	puts("-1");
    return 0;
}

 

The football season has just ended in Berland. According to the rules of Berland football, each match is played between two teams. The result of each match is either a draw, or a victory of one of the playing teams. If a team wins the match, it gets ww points, and the opposing team gets 00 points. If the game results in a draw, both teams get dd points.

The manager of the Berland capital team wants to summarize the results of the season, but, unfortunately, all information about the results of each match is lost. The manager only knows that the team has played nn games and got pp points for them.

You have to determine three integers xx, yy and zz — the number of wins, draws and loses of the team. If there are multiple answers, print any of them. If there is no suitable triple (x,y,z)(x,y,z), report about it.

Input

The first line contains four integers nn, pp, ww and dd (1≤n≤1012,0≤p≤1017,1≤ddw>d, so the number of points awarded for winning is strictly greater than the number of points awarded for draw.

Output

If there is no answer, print −1−1.

Otherwise print three non-negative integers xx, yy and zz — the number of wins, draws and losses of the team. If there are multiple possible triples (x,y,z)(x,y,z), print any of them. The numbers should meet the following conditions:

• x⋅w+y⋅d=px⋅w+y⋅d=p,
• x+y+z=nx+y+z=n.

Examples

input

Copy

30 60 3 1

output

Copy

17 9 4

input

Copy

10 51 5 4

output

Copy

-1

input

Copy

20 0 15 5

output

Copy

0 0 20

Note

One of the possible answers in the first example — 1717 wins, 99 draws and 44 losses. Then the team got 17⋅3+9⋅1=6017⋅3+9⋅1=60 points in 17+9+4=3017+9+4=30 games.

In the second example the maximum possible score is 10⋅5=5010⋅5=50. Since p=51p=51, there is no answer.

In the third example the team got 00 points, so all 2020 games were lost.