Problem description

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

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

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

5 1
BGGBG

GBGGB

5 2
BGGBG

GGBGB

4 1
GGGB

GGGB
解題思路：題目的意思就是輸入一個字符串（只由‘B‘和‘G‘兩個字符組成）和變換時間t秒。要求每秒都將當前的字符串中這兩個連續的字符‘B‘（在前）‘G‘（在後）交換位置，註意每秒中每個字符至多變換1次，水過！
AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int n,t;char s[55];
 5     cin>>n>>t;getchar();
 6     cin>>s;
 7     for(int i=1;i<=t;++i)//變換時間t秒
 8         for(int j=1;s[j]!=‘\0‘;++j)//每秒中每個字符至多變換1次，不能重復變字符
 9             if((s[j-1]==‘B‘)&&(s[j]==‘G‘)){swap(s[j-1],s[j]);j++;}//j++是跳過已變換的字符，避免變換兩次或多次
10     cout<<s<<endl;
11     return 0;
12 }

C - Queue at the School