【Queue at the School】【CodeForces - 266B】（模拟）

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我是靠谱客的博主含糊啤酒，最近开发中收集的这篇文章主要介绍【Queue at the School】【CodeForces - 266B】（模拟），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

题目：

During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.

Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.

You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.

Input

The first line contains two integers n and t (1 ≤ n, t ≤ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.

The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".

Output

Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".

Examples

Input

5 1
BGGBG

Output

GBGGB

Input

5 2
BGGBG

Output

GGBGB

Input

4 1
GGGB

Output

GGGB

解题思路：虽说是和队列有关，但实际上除了题目的讲解是有关的，貌似没有涉及过多，这个题目就是体现女士优先的原则，在每一秒如果有男士在女士前边，二者就应该换一下位置。这个会有一个时间期限，作为限制。

ac代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
const int maxn = 1e4+1000;

int main()
{
	int n,t;
	char str[1000];
	scanf("%d%d",&n,&t);
	cin>>str;
	while(t--)
	{
		for(int i=0;i<n-1;i++)
		{
			if(str[i]=='B'&&str[i+1]=='G')
			{
				str[i]='G';
				str[i+1]='B';
				i++;
			}
		}
	}
	cout<<str;
}