CF1244C.The Football Season

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 $w$ points, and the opposing team gets $0$ points. If the game results in a draw, both teams get $d$ 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 $n$ games and got $p$ points for them.

You have to determine three integers $x$, $y$ and $z$ — 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 $(x, y, z)$, report about it.

贝兰德足球赛季刚刚结束。根据贝兰德足球规则，每场比赛在两支队伍之间进行。每场比赛的结果要么是平局，要么是一方获胜。若一支队伍获胜，则获得 $w$ 分，而对手获得 $0$ 分；若比赛以平局结束，则双方各得 $d$ 分。

贝兰德首都队的教练希望总结本赛季的成绩，但不幸的是，所有关于各场比赛结果的信息均已丢失。教练仅知道该队共进行了 $n$ 场比赛，并总共获得了 $p$ 分。

你需要确定三个整数 $x$、$y$ 和 $z$ —— 分别表示该队的胜场数、平局数和负场数。若存在多组解，输出任意一组即可；若不存在满足条件的三元组 $(x, y, z)$，则报告无解。

The first line contains four integers $n$, $p$, $w$ and $d$ $(1 \le n \le 10^{12}, 0 \le p \le 10^{17}, 1 \le d \lt w \le 10^{5})$ — the number of games, the number of points the team got, the number of points awarded for winning a match, and the number of points awarded for a draw, respectively. Note that $w \gt d$, so the number of points awarded for winning is strictly greater than the number of points awarded for draw.

第一行包含四个整数 $n$、$p$、$w$ 和 $d$ $(1 \le n \le 10^{12}, 0 \le p \le 10^{17}, 1 \le d \lt w \le 10^{5})$，分别表示比赛场数、该队获得的总分数、获胜一场所得分数以及平局一场所得分数。注意 $w \gt d$，即获胜所得分数严格大于平局所得分数。

If there is no answer, print $-1$.

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

• $x \cdot w + y \cdot d = p$,
• $x + y + z = n$.

如果没有答案，输出 $-1$。

否则，输出三个非负整数 $x$、$y$ 和 $z$ —— 分别表示该队的胜场数、平场数和负场数。如果存在多个可能的三元组 $(x, y, z)$，输出其中任意一个即可。这些数字需满足以下条件：

• $x \cdot w + y \cdot d = p$，
• $x + y + z = n$。

One of the possible answers in the first example — $17$ wins, $9$ draws and $4$ losses. Then the team got $17 \cdot 3 + 9 \cdot 1 = 60$ points in $17 + 9 + 4 = 30$ games.

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

In the third example the team got $0$ points, so all $20$ games were lost.

第一个样例的一个可能答案是：$17$ 场胜利、$9$ 场平局和 $4$ 场失利。此时该队在 $17 + 9 + 4 = 30$ 场比赛中共获得 $17 \cdot 3 + 9 \cdot 1 = 60$ 分。

第二个样例中，最高可能得分是 $10 \cdot 5 = 50$。由于 $p = 51$，因此无解。

第三个样例中，该队得分为 $0$，因此全部 $20$ 场比赛均告失利。