題幹：

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

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

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

題目大意：

輸入n（學生人數），t（時間），每秒切換一次位置，為了體現女士優先的紳士風格，站在佇列靠前的男生與緊鄰靠後的女生交換一次位置，每秒交換一次（一趟下來所有符合條件的都交換一次），求t 秒後，佇列分佈情況。

解題報告：

就直接模擬就行了。

AC程式碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
char s[55];
int main() {
	int n,t;
	scanf("%d%d\n",&n,&t);
	scanf("%s",s);
	while(t--) {
		for (int i=0;i<n;) {
			if(s[i]=='B') {
				if(s[i+1]=='G') {
					swap(s[i],s[i+1]);
					i+=2;
				} 
				else i++;
			} 
			else i++;
		}
	}
	printf("%s\n",s);
	return 0;
}