#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') f = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = (x << 1) + (x << 3) + (op ^ 48);
    op = getchar();
  }
  return x * f;
}
int n, m, a[100][100], b[100][100], c[100][100], ans;
int main() {
  n = read();
  m = read();
  if (n < 2 || m < 2) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = read();
  for (int i = 1; i < n; ++i)
    for (int j = 1; j < m; ++j) {
      if (a[i][j] && !b[i][j]) {
        ans++;
        c[i][j] = 1;
        if (!a[i + 1][j + 1]) {
          cout << -1;
          return 0;
        } else
          b[i + 1][j + 1] = 1;
        if (!a[i + 1][j]) {
          cout << -1;
          return 0;
        } else
          b[i + 1][j] = 1;
        if (!a[i][j + 1]) {
          cout << -1;
          return 0;
        } else
          b[i][j + 1] = 1;
      }
      if (a[i][j] && b[i][j]) {
        if (a[i + 1][j + 1] && a[i][j + 1] && a[i + 1][j])
          ans++, b[i + 1][j + 1] = b[i][j + 1] = b[i + 1][j] = c[i][j] = 1;
      }
    }
  for (int i = 1; i <= n; ++i)
    if (b[i][m] != a[i][m]) {
      cout << -1;
      return 0;
    }
  for (int i = 1; i <= m; ++i)
    if (b[n][i] != a[n][i]) {
      cout << -1;
      return 0;
    }
  cout << ans << endl;
  for (int i = 1; i < n; ++i)
    for (int j = 1; j < m; ++j)
      if (c[i][j]) cout << i << ' ' << j << endl;
}

### Prompt

Please formulate 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;
inline int read() {
  int x = 0, f = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') f = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = (x << 1) + (x << 3) + (op ^ 48);
    op = getchar();
  }
  return x * f;
}
int n, m, a[100][100], b[100][100], c[100][100], ans;
int main() {
  n = read();
  m = read();
  if (n < 2 || m < 2) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = read();
  for (int i = 1; i < n; ++i)
    for (int j = 1; j < m; ++j) {
      if (a[i][j] && !b[i][j]) {
        ans++;
        c[i][j] = 1;
        if (!a[i + 1][j + 1]) {
          cout << -1;
          return 0;
        } else
          b[i + 1][j + 1] = 1;
        if (!a[i + 1][j]) {
          cout << -1;
          return 0;
        } else
          b[i + 1][j] = 1;
        if (!a[i][j + 1]) {
          cout << -1;
          return 0;
        } else
          b[i][j + 1] = 1;
      }
      if (a[i][j] && b[i][j]) {
        if (a[i + 1][j + 1] && a[i][j + 1] && a[i + 1][j])
          ans++, b[i + 1][j + 1] = b[i][j + 1] = b[i + 1][j] = c[i][j] = 1;
      }
    }
  for (int i = 1; i <= n; ++i)
    if (b[i][m] != a[i][m]) {
      cout << -1;
      return 0;
    }
  for (int i = 1; i <= m; ++i)
    if (b[n][i] != a[n][i]) {
      cout << -1;
      return 0;
    }
  cout << ans << endl;
  for (int i = 1; i < n; ++i)
    for (int j = 1; j < m; ++j)
      if (c[i][j]) cout << i << ' ' << j << endl;
}

```