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 long long int mod = 1e9 + 7;
const int OO = INT_MAX, P = 200;
int xd[] = {1, 0, -1, 0, -1, 1, 1, -1};
int yd[] = {0, -1, 0, 1, 1, 1, -1, -1};
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long int n, m, sb = 0;
  char arr[P][P];
  cin >> n >> m;
  pair<long long int, long long int> s;
  pair<long long int, long long int> f;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      cin >> arr[i][j];
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      if (sb == 0 && arr[i][j] == 'B') {
        sb = 1;
        s.first = i, s.second = j;
        f.first = i, f.second = j;
      } else if (arr[i][j] == 'B') {
        f.first = i, f.second = j;
      }
    }
  }
  f.first -= s.first, f.second -= s.second;
  f.first /= 2, f.second /= 2;
  f.first += s.first, f.second += s.second;
  f.first++, f.second++;
  cout << f.first << " " << f.second;
  return 0;
}
