human
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
Using python3 can you solve the prior task?
assistant
#Here is Python code to determine the way the queue is going to look after t seconds.

import os
import gc
import sys
from io import BytesIO, IOBase
from collections import Counter
from collections import deque
import heapq
import math
import statistics

def sin():
    return input()

def ain():
    return list(map(int, sin().split()))

def sain():
    return input().split()

def iin():
    return int(sin())

MAX = float('inf')
MIN = float('-inf')
MOD = 1000000007

def readTree(n, m):
    adj = [deque([]) for _ in range(n+1)]
    for _ in range(m):
        u,v = ain()
        adj[u].append(v)
        adj[v].append(u)
    return adj

def solve(n, d):
    pass

def main():
    n, k = ain()
    l = list(sin())
    for i in range(k):
        j = 0
        while j < (n-1):
            if l[j] == 'B' and l[j+1] == 'G':
                l[j], l[j+1] = l[j+1], l[j]
                j += 2
            else:
                j+=1
    print(''.join(l))

        
        
        

    








# Fast IO Template starts

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
         ...