#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m;
int vis[N], l[N], r[N];
char s[N], ans[N];
queue<int> q;
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  memset(vis, 255, sizeof(vis));
  for (int i = 1; i <= n; ++i) l[i] = i - 1, r[i] = i + 1;
  l[1] = n;
  r[n] = 1;
  for (int i = 1; i <= n; ++i)
    if (s[l[i]] == s[i] || s[r[i]] == s[i]) {
      vis[i] = 0;
      q.push(i);
    }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (vis[l[u]] == -1) {
      vis[l[u]] = vis[u] + 1;
      q.push(l[u]);
    }
    if (vis[r[u]] == -1) {
      vis[r[u]] = vis[u] + 1;
      q.push(r[u]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] > m || vis[i] == -1) {
      if (m & 1)
        printf("%c", 'W' + 'B' - s[i]);
      else
        printf("%c", s[i]);
    } else {
      if (vis[i] & 1)
        printf("%c", 'W' + 'B' - s[i]);
      else
        printf("%c", s[i]);
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

There are n chips arranged in a circle, numbered from 1 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 2 to (n - 1) two neighbouring chips have numbers (i - 1) and (i + 1). The neighbours for the chip i = 1 are n and 2. The neighbours of i = n are (n - 1) and 1.

The following picture describes one iteration with n = 6. The chips 1, 3 and 4 are initially black, and the chips 2, 5 and 6 are white. After the iteration 2, 3 and 4 become black, and 1, 5 and 6 become white.

<image>

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

Input

The first line contains two integers n and k (3 ≤ n ≤ 200 000, 1 ≤ k ≤ 10^{9}) — 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


6 1
BWBBWW


Output


WBBBWW


Input


7 3
WBWBWBW


Output


WWWWWWW


Input


6 4
BWBWBW


Output


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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m;
int vis[N], l[N], r[N];
char s[N], ans[N];
queue<int> q;
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  memset(vis, 255, sizeof(vis));
  for (int i = 1; i <= n; ++i) l[i] = i - 1, r[i] = i + 1;
  l[1] = n;
  r[n] = 1;
  for (int i = 1; i <= n; ++i)
    if (s[l[i]] == s[i] || s[r[i]] == s[i]) {
      vis[i] = 0;
      q.push(i);
    }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (vis[l[u]] == -1) {
      vis[l[u]] = vis[u] + 1;
      q.push(l[u]);
    }
    if (vis[r[u]] == -1) {
      vis[r[u]] = vis[u] + 1;
      q.push(r[u]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] > m || vis[i] == -1) {
      if (m & 1)
        printf("%c", 'W' + 'B' - s[i]);
      else
        printf("%c", s[i]);
    } else {
      if (vis[i] & 1)
        printf("%c", 'W' + 'B' - s[i]);
      else
        printf("%c", s[i]);
    }
  }
  printf("\n");
  return 0;
}

```