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 maxn = 1050;
char mp[maxn][maxn];
int dp[maxn][maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> mp[i][j];
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (mp[i][j] == 'B') {
        cnt++;
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + 1;
      } else
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
    }
  int bei = 0, bej = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (mp[i][j] == 'W') continue;
      if (dp[i][j] == 1) bei = i, bej = j;
      if (dp[i][j] == cnt) {
        printf("%d %d\n", (i + bei) / 2, (j + bej) / 2);
        return 0;
      }
    }
  return 0;
}
