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 comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void pre() {}
void solve() {}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pre();
  int n, m;
  cin >> n >> m;
  int fx, fy, lx, ly;
  fx = -1, fy = -1;
  string s[n];
  for (int i = 0; i < (n); ++i) cin >> s[i];
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j) {
      if (s[i][j] == 'B') {
        if (fx == -1) fx = i, fy = j;
        lx = i, ly = j;
      }
    }
  cout << (fx + lx) / 2 + 1 << ' ' << (fy + ly) / 2 + 1;
  return 0;
}
