#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 1488;
string s;
long long n, k, chl = 1;
string one() {
  string s2 = s;
  chl = 0;
  for (long long i = 0; i < n; i++) {
    long long e0 = s[(i - 1 + n) % n] == 'W' ? 1 : 0;
    long long e1 = s[i] == 'W' ? 1 : 0;
    long long e2 = s[(i + 1) % n] == 'W' ? 1 : 0;
    if (e0 + e1 + e2 >= 2)
      s2[i] = 'W';
    else
      s2[i] = 'B';
    if (s[i] != s2[i]) chl++;
  }
  return s2;
}
bool odin[N];
long long ps[N];
bool ban[N];
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  cin >> s;
  long long oc = 0;
  for (long long i = 0; i < n; i++) {
    long long e0 = s[(i - 1 + n) % n] == 'W' ? 1 : 0;
    long long e1 = s[i] == 'W' ? 1 : 0;
    long long e2 = s[(i + 1) % n] == 'W' ? 1 : 0;
    if (e1 != e0 && e1 != e2) odin[i] = 1, oc++;
  }
  if (oc == n) {
    if (k & 1) s = one();
    cout << s << '\n';
    return 0;
  }
  long long wp = n - 1;
  while (odin[wp]) --wp;
  for (long long i = (wp + 1) % n, j = 0; j < n; j++, i = (i + 1) % n) {
    if (odin[i])
      ps[i] = ps[(i + n - 1) % n] + 1;
    else
      ps[i] = 0;
  }
  wp = 0;
  while (odin[wp]) ++wp;
  for (long long i = (wp - 1 + n) % n, j = 0; j < n; j++, i = (i - 1 + n) % n) {
    if (odin[i]) ps[i] = min(ps[i], ps[(i + 1) % n] + 1);
  }
  long long p2 = wp;
  ban[wp] = 1;
  while (true) {
    bool cd = 1;
    while (!odin[p2]) {
      p2 = (p2 + 1) % n;
      if (ban[p2]) {
        cd = 0;
        break;
      }
      ban[p2] = 1;
    }
    if (!cd) break;
    long long p1 = p2;
    while (odin[p2]) {
      p2 = (p2 + 1) % n;
      if (ban[p2] && odin[p2]) {
        cd = 0;
        break;
      }
      ban[p2] = 1;
    }
    if (!cd) break;
    long long p3 = p2;
    p2 = (p2 + n - 1) % n;
    long long us = (p1 <= p2 ? p2 - p1 + 1 : n - p1 + p2 + 1);
    for (long long op = 0; op < k && us > 0; op++) {
      odin[p1] = 0, odin[p2] = 0;
      if (p1 <= p2) {
        s[p1] = s[(p1 + n - 1) % n];
        s[p2] = s[(p2 + 1) % n];
        ++p1, p2--;
        us -= 2;
      } else {
        s[p1] = s[(p1 + n - 1) % n];
        s[p2] = s[(p2 + 1) % n];
        p1 = (p1 + 1) % n;
        p2 = (p2 + n - 1) % n;
        us -= 2;
      }
    }
    p2 = p3;
  }
  for (long long i = 0; i < n; i++) {
    if (odin[i] && k % 2 == 1) {
      s[i] = (s[i] == 'W' ? 'B' : 'W');
    }
  }
  cout << s;
}

### 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;
const long long N = 2e5 + 1488;
string s;
long long n, k, chl = 1;
string one() {
  string s2 = s;
  chl = 0;
  for (long long i = 0; i < n; i++) {
    long long e0 = s[(i - 1 + n) % n] == 'W' ? 1 : 0;
    long long e1 = s[i] == 'W' ? 1 : 0;
    long long e2 = s[(i + 1) % n] == 'W' ? 1 : 0;
    if (e0 + e1 + e2 >= 2)
      s2[i] = 'W';
    else
      s2[i] = 'B';
    if (s[i] != s2[i]) chl++;
  }
  return s2;
}
bool odin[N];
long long ps[N];
bool ban[N];
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  cin >> s;
  long long oc = 0;
  for (long long i = 0; i < n; i++) {
    long long e0 = s[(i - 1 + n) % n] == 'W' ? 1 : 0;
    long long e1 = s[i] == 'W' ? 1 : 0;
    long long e2 = s[(i + 1) % n] == 'W' ? 1 : 0;
    if (e1 != e0 && e1 != e2) odin[i] = 1, oc++;
  }
  if (oc == n) {
    if (k & 1) s = one();
    cout << s << '\n';
    return 0;
  }
  long long wp = n - 1;
  while (odin[wp]) --wp;
  for (long long i = (wp + 1) % n, j = 0; j < n; j++, i = (i + 1) % n) {
    if (odin[i])
      ps[i] = ps[(i + n - 1) % n] + 1;
    else
      ps[i] = 0;
  }
  wp = 0;
  while (odin[wp]) ++wp;
  for (long long i = (wp - 1 + n) % n, j = 0; j < n; j++, i = (i - 1 + n) % n) {
    if (odin[i]) ps[i] = min(ps[i], ps[(i + 1) % n] + 1);
  }
  long long p2 = wp;
  ban[wp] = 1;
  while (true) {
    bool cd = 1;
    while (!odin[p2]) {
      p2 = (p2 + 1) % n;
      if (ban[p2]) {
        cd = 0;
        break;
      }
      ban[p2] = 1;
    }
    if (!cd) break;
    long long p1 = p2;
    while (odin[p2]) {
      p2 = (p2 + 1) % n;
      if (ban[p2] && odin[p2]) {
        cd = 0;
        break;
      }
      ban[p2] = 1;
    }
    if (!cd) break;
    long long p3 = p2;
    p2 = (p2 + n - 1) % n;
    long long us = (p1 <= p2 ? p2 - p1 + 1 : n - p1 + p2 + 1);
    for (long long op = 0; op < k && us > 0; op++) {
      odin[p1] = 0, odin[p2] = 0;
      if (p1 <= p2) {
        s[p1] = s[(p1 + n - 1) % n];
        s[p2] = s[(p2 + 1) % n];
        ++p1, p2--;
        us -= 2;
      } else {
        s[p1] = s[(p1 + n - 1) % n];
        s[p2] = s[(p2 + 1) % n];
        p1 = (p1 + 1) % n;
        p2 = (p2 + n - 1) % n;
        us -= 2;
      }
    }
    p2 = p3;
  }
  for (long long i = 0; i < n; i++) {
    if (odin[i] && k % 2 == 1) {
      s[i] = (s[i] == 'W' ? 'B' : 'W');
    }
  }
  cout << s;
}

```