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;
using i64 = long long;
using f64 = double;
using pii = pair<int, int>;
using pll = pair<i64, i64>;
const int N = 120;
char mx[N][N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m, l = 0, r = 0, u = 0, d = 0;
  bool flag;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> (mx[i] + 1);
  flag = true;
  for (int i = 1; i <= n && flag; ++i)
    for (int j = 1; j <= m && flag; ++j)
      if (mx[i][j] == 'B') {
        l = j;
        u = i;
        flag = false;
      }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (mx[i][j] == 'B') {
        r = max(r, j);
        d = max(d, i);
      }
  cout << (u + d) / 2 << ' ' << (l + r) / 2 << endl;
  return 0;
}
