F. Chips

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are ?n chips arranged in a circle, numbered from 11 to ?n.

Initially each chip has black or white color. Then ?k iterations occur. During each iteration the chips change their colors according to the following rules. For each chip ?i, three chips are considered: chip ?i itself and two its neighbours. If the number of white chips among these three is greater than the number of black chips among these three chips, then the chip ?i becomes white. Otherwise, the chip ?i becomes black.

Note that for each ?i from 22 to (?−1)(n−1) two neighbouring chips have numbers (?−1)(i−1) and (?+1)(i+1). The neighbours for the chip ?=1i=1 are ?n and 22. The neighbours of ?=?i=n are (?−1)(n−1) and 11.

The following picture describes one iteration with ?=6n=6. The chips 11, 33 and 44 are initially black, and the chips 22, 55 and 66 are white. After the iteration 22, 33 and 44 become black, and 11, 55 and 66 become white.

1244F_ios

Your task is to determine the color of each chip after ?k iterations.

Input

The first line contains two integers ?n and ?k (3≤?≤200000,1≤?≤109)(3≤n≤200000,1≤k≤109) — the number of chips and the number of iterations, respectively.

The second line contains a string consisting of ?n characters "W" and "B". If the ?i-th character is "W", then the ?i-th chip is white initially. If the ?i-th character is "B", then the ?i-th chip is black initially.

Output

Print a string consisting of ?n characters "W" and "B". If after ?k iterations the ?i-th chip is white, then the ?i-th character should be "W". Otherwise the ?i-th character should be "B".

Examples

input

Copy

6 1
BWBBWW

output

Copy

WBBBWW

input

Copy

7 3
WBWBWBW

output

Copy

WWWWWWW

input

Copy

6 4
BWBWBW

output

Copy

BWBWBW

Note

The first example is described in the statement.

The second example: "WBWBWBW" →→ "WWBWBWW" →→ "WWWBWWW" →→ "WWWWWWW". So all chips become white.

The third example: "BWBWBW" →→ "WBWBWB" →→ "BWBWBW" →→ "WBWBWB" →→ "BWBWBW".

题解：

直接模拟的时间复杂度是O(nk)。

寻找规律，如果有两点连续且同色，那么是肯定不会变色的，然后每次变换会向左右扩展一次，并使用visit[i]来计算需要多少次变换才能扩展到该点。

如果该点与左右都不同色那么会交替变色：BWBW->WBWB->BWBW....

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <cstdlib>
#include <stack>
using namespace std;

int main(){
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;

  vector<int> visit(n, -1);
  queue<int> q;
  for (int i = 0; i < n; i++) {
    int left = (i + n - 1) % n;
    int right = (i + 1) % n;
    if (s[i] != s[left] && s[i] != s[right]) {
      continue;
    }
    else {
      visit[i] = 0;
      q.push(i);
    }
  }
  while (q.empty() == false) {
    int idx = q.front();
    q.pop();
    int left = (idx + n - 1) % n;
    int right = (idx + 1) % n;
    if (visit[left] == -1) {
      visit[left] = visit[idx] + 1;
      q.push(left);
    }
    if (visit[right] == -1) {
      visit[right] = visit[idx] + 1;
      q.push(right);
    }
  }
  for (int i = 0; i < n; i++) {
    if (visit[i] == -1 || visit[i] > k) {
      if (k % 2 == 1) {
        s[i] = 'B' + 'W' - s[i];
      }
    }
    else {
      if (visit[i] % 2 == 1) {
        s[i] = 'B' + 'W' - s[i];
      }
    }
  }
  cout << s << endl;
  return 0;
}