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 INF = (int)2e9;
const double EPS = (double)1e-9;
const double PI = (double)acos(-1);
const long long MOD = (long long)1e9 + 7;
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  gcd(a % b, b);
}
bool palin(string s) {
  int len = s.length();
  for (int i = 0; i < len / 2; i++) {
    if (s[i] != s[len - 1 - i]) {
      return false;
    }
  }
  return true;
}
void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  int n, m, a, b, cnt = -1, done = 0;
  string s[120];
  cin >> n >> m;
  for (int i = 1; i <= n; i += 1) cin >> s[i];
  for (int i = 1; i <= n; i += 1) {
    for (int j = 0; j < m; j++) {
      if (s[i][j] == 'B') {
        a = i;
        b = j;
        for (;;) {
          cnt++;
          if (s[i + cnt][j + cnt] != 'B') break;
        }
        done = 1;
        break;
      }
    }
    if (done) break;
  }
  cout << a + (cnt / 2) << " " << b + 1 + (cnt / 2) << "\n";
  return 0;
}
