Queue at the School

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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 
   
复制代码5 1
BGGBG
5 1
BGGBG

     output 
   
复制代码GBGGB
GBGGB

     input 
   
复制代码5 2
BGGBG
5 2
BGGBG

     output 
   
复制代码GGBGB
GGBGB

     input 
   
复制代码4 1
GGGB
4 1
GGGB

     output 
   
复制代码GGGBGGGB

有n个同学在餐厅排队，如果男同学在前面就会给后面紧挨着的女同学让个位置，但是对于一列而言只能同时操作t次；问t次后的队列应该什么样子的

复制代码

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#include <stdio.h>
#include <string.h>
int main()
{
	int n,t,i,j;
	char a[50],p;
	scanf ("%d%d",&n,&t);
	scanf ("%s",a);
	a[n]='';
	for (j=0;j<t;j++)//进行t次交换 
	{
	for (i=0;i<n-1;i++)
	{
		if (a[i]=='B'&&a[i+1]=='G') //判断是否需要交换 
		{
			p=a[i];
			a[i]=a[i+1];
			a[i+1]=p;
			i++;
		}
	}
    }
    puts(a);
}