#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, t;
  scanf("%d %d", &n, &m);
  bool a[n][m], fail = false;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &t);
      if (t == 0) {
        a[i][j] = false;
      } else {
        a[i][j] = true;
      }
    }
  }
  vector<int> x, y;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!a[i][j]) {
        continue;
      }
      bool aaa = false;
      for (int k = -1; k < 1; k++) {
        for (int l = -1; l < 1; l++) {
          if (i + k < 0 || i + k + 1 >= n || j + l < 0 || j + l + 1 >= m ||
              aaa) {
            continue;
          }
          bool b = a[i + k][j + l] && a[i + k][j + l + 1] &&
                   a[i + k + 1][j + l] && a[i + k + 1][j + l + 1];
          if (b) {
            x.push_back(i + k);
            y.push_back(j + l);
            aaa = true;
          }
        }
      }
      if (!aaa) {
        fail = true;
      }
    }
  }
  if (fail) {
    printf("-1\n");
  } else {
    printf("%d\n", x.size());
    for (int i = 0; i < x.size(); i++) {
      printf("%d %d\n", x[i] + 1, y[i] + 1);
    }
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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} 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, t;
  scanf("%d %d", &n, &m);
  bool a[n][m], fail = false;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &t);
      if (t == 0) {
        a[i][j] = false;
      } else {
        a[i][j] = true;
      }
    }
  }
  vector<int> x, y;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!a[i][j]) {
        continue;
      }
      bool aaa = false;
      for (int k = -1; k < 1; k++) {
        for (int l = -1; l < 1; l++) {
          if (i + k < 0 || i + k + 1 >= n || j + l < 0 || j + l + 1 >= m ||
              aaa) {
            continue;
          }
          bool b = a[i + k][j + l] && a[i + k][j + l + 1] &&
                   a[i + k + 1][j + l] && a[i + k + 1][j + l + 1];
          if (b) {
            x.push_back(i + k);
            y.push_back(j + l);
            aaa = true;
          }
        }
      }
      if (!aaa) {
        fail = true;
      }
    }
  }
  if (fail) {
    printf("-1\n");
  } else {
    printf("%d\n", x.size());
    for (int i = 0; i < x.size(); i++) {
      printf("%d %d\n", x[i] + 1, y[i] + 1);
    }
  }
}

```