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 N = 116;
char s[N][N];
int a[N][N], n, m;
int sum[N][N];
int stx, sty, enx, eny;
int nowmax;
bool vis;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i] + 1);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (s[i][j] == 'W')
        a[i][j] = 0;
      else
        a[i][j] = 1;
      sum[i][j] = a[i][j] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
      if (sum[i][j] == 1 && !vis) stx = i, sty = j, vis = 1;
      if (sum[i][j] && sum[i][j] > nowmax) enx = i, eny = j;
      nowmax = max(nowmax, sum[i][j]);
    }
  }
  cout << (stx + enx) / 2 << " " << (sty + eny) / 2 << endl;
  return 0;
}
