#include <bits/stdc++.h>
using namespace std;
template <class T>
void read(T &x) {
  char c;
  bool op = 0;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') op = 1;
  x = c - '0';
  while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
  if (op) x = -x;
}
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
const int maxn = 2e5 + 5;
char s[maxn];
int color[maxn * 3];
struct Segment {
  int l, r, color;
} segment[maxn * 3];
int n, k, tot;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  cin >> (s + 1);
  for (int i = 1; i <= n; i++) color[i] = s[i] == 'B' ? 1 : 0;
  for (int i = n + 1; i <= 3 * n; i++) color[i] = color[(i - 1) % n + 1];
  for (int i = 1, j = 1; i <= 3 * n; i = j + 1) {
    while (j + 1 <= 3 * n && color[i] == color[j + 1]) ++j;
    if (j - i >= 1) segment[++tot] = {i, j, color[i]};
  }
  for (int i = 1; i < tot; i++) {
    int l = segment[i].r + 1, r = segment[i + 1].l - 1;
    int lc = color[l - 1], rc = color[r + 1];
    if (l > r) continue;
    int len = r - l + 1;
    if (2 * k >= len) {
      if (lc == rc) {
        for (int j = l; j <= r; j++) color[j] = lc;
      } else {
        for (int j = l; j <= l + len / 2 - 1; j++) color[j] = lc;
        for (int j = r; j >= r - len / 2 + 1; j--) color[j] = rc;
      }
    } else {
      for (int j = l; j <= l + k - 1; j++) color[j] = lc;
      for (int j = r; j >= r - k + 1; j--) color[j] = rc;
      for (int j = l + k; j <= r - k; j++) color[j] = !color[j - 1];
    }
  }
  if (tot == 0 && k % 2)
    for (int i = n + 1; i <= 2 * n; i++) color[i] = !color[i - 1];
  for (int i = n + 1; i <= 2 * n; i++) printf("%c", color[i] ? 'B' : 'W');
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
template <class T>
void read(T &x) {
  char c;
  bool op = 0;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') op = 1;
  x = c - '0';
  while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
  if (op) x = -x;
}
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
const int maxn = 2e5 + 5;
char s[maxn];
int color[maxn * 3];
struct Segment {
  int l, r, color;
} segment[maxn * 3];
int n, k, tot;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  cin >> (s + 1);
  for (int i = 1; i <= n; i++) color[i] = s[i] == 'B' ? 1 : 0;
  for (int i = n + 1; i <= 3 * n; i++) color[i] = color[(i - 1) % n + 1];
  for (int i = 1, j = 1; i <= 3 * n; i = j + 1) {
    while (j + 1 <= 3 * n && color[i] == color[j + 1]) ++j;
    if (j - i >= 1) segment[++tot] = {i, j, color[i]};
  }
  for (int i = 1; i < tot; i++) {
    int l = segment[i].r + 1, r = segment[i + 1].l - 1;
    int lc = color[l - 1], rc = color[r + 1];
    if (l > r) continue;
    int len = r - l + 1;
    if (2 * k >= len) {
      if (lc == rc) {
        for (int j = l; j <= r; j++) color[j] = lc;
      } else {
        for (int j = l; j <= l + len / 2 - 1; j++) color[j] = lc;
        for (int j = r; j >= r - len / 2 + 1; j--) color[j] = rc;
      }
    } else {
      for (int j = l; j <= l + k - 1; j++) color[j] = lc;
      for (int j = r; j >= r - k + 1; j--) color[j] = rc;
      for (int j = l + k; j <= r - k; j++) color[j] = !color[j - 1];
    }
  }
  if (tot == 0 && k % 2)
    for (int i = n + 1; i <= 2 * n; i++) color[i] = !color[i - 1];
  for (int i = n + 1; i <= 2 * n; i++) printf("%c", color[i] ? 'B' : 'W');
  return 0;
}

```