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 SD() {
  int x;
  scanf("%d", &x);
  return x;
}
long long SLL() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const int AKA = 5e2 + 5;
char s[AKA][AKA];
int fr[26];
int main() {
  int ACCEPTED = 1;
  ;
  while (ACCEPTED--) {
    int n = SD(), m = SD();
    int sc, sr, ec, er;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) cin >> s[i][j];
    }
    for (int i = 1; i <= n; ++i) {
      int counter = 0;
      bool f1 = false;
      for (int j = 1; j <= m; ++j)
        if (s[i][j] == 'B') {
          sr = j;
          while (s[i][j] == 'B') ++counter, ++j;
          er = counter - 1 + sr;
          f1 = true;
        }
      if (f1) break;
    }
    for (int i = 1; i <= n; ++i) {
      bool f1 = false;
      int counter = 0;
      for (int j = 1; j <= m; ++j)
        if (s[j][i] == 'B') {
          sc = i;
          while (s[j][i] == 'B') ++counter, ++i;
          ec = counter - 1 + sc;
          f1 = true;
        }
      if (f1) break;
    }
    for (int j = 1; j <= m; ++j) {
      bool f1 = false;
      int counter = 0;
      for (int i = 1; i <= n; ++i)
        if (s[i][j] == 'B') {
          sc = i;
          while (s[i][j] == 'B') ++counter, ++i;
          ec = counter - 1 + sc;
          f1 = true;
        }
      if (f1) break;
    }
    printf("%d %d", (sc + ec) / 2, (sr + er) / 2);
  }
}
