Consider a table of size n × m, initially fully white. Rows are numbered 1 through n from top to bottom, columns 1 through m from left to right. Some square inside the table with odd side length was painted black. Find the center of this square.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 115) — the number of rows and the number of columns in the table.

The i-th of the next n lines contains a string of m characters s_{i1} s_{i2} … s_{im} (s_{ij} is 'W' for white cells and 'B' for black cells), describing the i-th row of the table.

Output

Output two integers r and c (1 ≤ r ≤ n, 1 ≤ c ≤ m) separated by a space — the row and column numbers of the center of the black square.

Examples

Input

5 6
WWBBBW
WWBBBW
WWBBBW
WWWWWW
WWWWWW


Output

2 4


Input

3 3
WWW
BWW
WWW


Output

2 1
#include <bits/stdc++.h>
using namespace std;
int a[150][150];
int g[150][150];
int m, n;
int get(int x1, int y1, int x2, int y2) {
  return g[x2][y2] + g[x1 - 1][y1 - 1] - g[x1 - 1][y2] - g[x2][y1 - 1];
}
int main() {
  if (fopen("a.txt", "r")) {
    freopen("a.txt", "r", stdin);
  } else
    ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> m >> n;
  for (int i = 1; i <= m; i++) {
    string s;
    cin >> s;
    for (int j = 1; j <= n; j++) a[i][j] = (s[j - 1] == 'B');
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++)
      g[i][j] = g[i][j - 1] + g[i - 1][j] - g[i - 1][j - 1] + a[i][j];
  }
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) {
      int limit = min(n - j + 1, m - i + 1);
      for (int r = limit; r >= 1; r--) {
        if (r % 2 == 0) continue;
        if (get(i, j, i + r - 1, j + r - 1) == (r * r)) {
          cout << i + r / 2 << ' ' << j + r / 2;
          return 0;
        }
      }
    }
}
