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;
inline int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long Long() {
  long x;
  scanf("%ld", &x);
  return x;
}
inline long long LLong() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline float Float() {
  float x;
  scanf("%f", &x);
  return x;
}
inline double Double() {
  double x;
  scanf("%lf", &x);
  return x;
}
inline pair<int, int> Pair() {
  pair<int, int> p;
  scanf("%d %d", &p.first, &p.second);
  return p;
}
void Char(char *x) { scanf("%s", x); }
void pI(int x) { printf("%d", x); }
void pL(long x) { printf("%ld", x); }
void pLL(long long x) { printf("%lld", x); }
void pUL(unsigned long long x) { printf("%llu", x); }
void pF(float x) { printf("%f", x); }
void pD(double x) { printf("%f", x); }
void pS(char x[]) { printf("%s", x); }
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9 + 7;
const int maxn = (int)2e5 + 5;
const int LOGN = 20;
char s[200][200];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, l = 5000, r = 0, u = 5000, d = 0;
  cin >> n >> m;
  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') {
        l = min(l, j + 1);
        r = max(r, j + 1);
        u = min(u, i + 1);
        d = max(d, i + 1);
      }
    }
  }
  cout << (u + d) / 2 << " " << (l + r) / 2;
  return 0;
}
