You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.

You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 × 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 ≤ x < n and 1 ≤ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.

Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.

Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50).

Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.

Output

If it is impossible to make B equal to A, print one integer -1.

Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k — the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 ≤ k ≤ 2500 should hold.

Examples

Input


3 3
1 1 1
1 1 1
0 1 1


Output


3
1 1
1 2
2 2


Input


3 3
1 0 1
1 0 1
0 0 0


Output


-1


Input


3 2
0 0
0 0
0 0


Output


0

Note

The sequence of operations in the first example:

\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & → & 1 & 1 & 0 & → & 1 & 1 & 1 & → & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} 
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5, inf = 1e18, mod = 1e9 + 7;
const long double pi = 3.141592653589793238;
bool check_prime(long long n) {
  long long flag = 0, i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (n == 1) {
    return false;
  } else if (flag == 0 || n == 2 || n == 3) {
    return true;
  } else {
    return false;
  }
}
long long fastexpo(long long a, long long b, long long MOD) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result = (result * a) % MOD;
    }
    a = (a * a) % MOD;
    b = b / 2;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, i, j, c, f;
  cin >> n >> m;
  long long a[n][m], b[n][m];
  c = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
      b[i][j] = 0;
      if (a[i][j] == 0) {
        c++;
      }
    }
  }
  if (c == n * m) {
    cout << "0\n";
  } else {
    vector<pair<long long, long long> > v;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (i + 1 < n && j + 1 < m) {
          if (a[i][j] == 1 && a[i][j + 1] == 1 && a[i + 1][j] == 1 &&
              a[i + 1][j + 1] == 1) {
            pair<long long, long long> pi;
            pi = {i + 1, j + 1};
            v.push_back(pi);
            b[i][j] = 1;
            b[i + 1][j + 1] = 1;
            b[i + 1][j] = 1;
            b[i][j + 1] = 1;
          }
        }
      }
    }
    f = 1;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (a[i][j] != b[i][j]) {
          f = 0;
          break;
        }
      }
    }
    if (f == 1) {
      cout << ((long long)((v).size())) << "\n";
      for (i = 0; i < ((long long)((v).size())); i++) {
        cout << v[i].first << " " << v[i].second << "\n";
      }
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}
