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;
const int N = 100005;
int dr8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dc8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dr4[] = {0, 1, -1, 0};
int dc4[] = {1, 0, 0, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  string a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int f = 0;
  int l = 0, st = 0, en = 0;
  for (int i = 0; i < n; i++) {
    if (f == 1) break;
    for (int j = 0; j < m; j++) {
      if (f == 1) break;
      if (a[i][j] == 'B') {
        l = 1;
        st = j;
        en = i;
        f = 1;
        for (int k = j + 1; k < m; k++) {
          if (a[i][k] == 'B')
            l++;
          else
            break;
        }
      }
    }
  }
  int k1 = (en + l - 1);
  int k2 = (st + l - 1);
  int an1 = (en + k1) / 2;
  int an2 = (st + k2) / 2;
  cout << an1 + 1 << " " << an2 + 1 << "\n";
}
