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>
#pragma GCC optimize("O2")
using namespace std;
const int MAX = 1e5 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const double pi = acos(-1);
const double eps = 1e-9;
int n, m, r, c, a, b;
bool st;
char x[150][150];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> x[i][j];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (x[i][j] == 'B') c++, st = 1;
    }
    if (st) break;
  }
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      if (x[i][j] == 'B' && !r) a = i, b = j;
      if (x[i][j] == 'B') r++, st = 0;
    }
    if (!st) break;
  }
  cout << a + r / 2 << " " << b + c / 2;
  return 0;
}
