#include <bits/stdc++.h>
using namespace std;
const int N = 60;
vector<int> v;
int n, m, a[N][N], b[N][N];
int main() {
  int kq = 0;
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
      if (a[i][j] == 1) kq++;
      b[i][j] = 0;
    }
  if (kq == 0) {
    cout << 0;
    return 0;
  }
  kq = 0;
  for (int i = 1; i <= n - 1; i++)
    for (int j = 1; j <= m - 1; j++) {
      if (a[i][j] == 1) {
        if (a[i + 1][j] == 1 && a[i][j + 1] == 1 && a[i + 1][j + 1] == 1) {
          b[i][j] = b[i + 1][j] = b[i][j + 1] = b[i + 1][j + 1] = 1;
          kq++;
          v.push_back(i);
          v.push_back(j);
        }
      }
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (a[i][j] != b[i][j]) {
        cout << -1;
        return 0;
      }
    }
  cout << kq << endl;
  for (int i = 0; i <= v.size() - 1; i += 2)
    cout << v[i] << " " << v[i + 1] << endl;
  return 0;
}

### 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;
const int N = 60;
vector<int> v;
int n, m, a[N][N], b[N][N];
int main() {
  int kq = 0;
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
      if (a[i][j] == 1) kq++;
      b[i][j] = 0;
    }
  if (kq == 0) {
    cout << 0;
    return 0;
  }
  kq = 0;
  for (int i = 1; i <= n - 1; i++)
    for (int j = 1; j <= m - 1; j++) {
      if (a[i][j] == 1) {
        if (a[i + 1][j] == 1 && a[i][j + 1] == 1 && a[i + 1][j + 1] == 1) {
          b[i][j] = b[i + 1][j] = b[i][j + 1] = b[i + 1][j + 1] = 1;
          kq++;
          v.push_back(i);
          v.push_back(j);
        }
      }
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (a[i][j] != b[i][j]) {
        cout << -1;
        return 0;
      }
    }
  cout << kq << endl;
  for (int i = 0; i <= v.size() - 1; i += 2)
    cout << v[i] << " " << v[i + 1] << endl;
  return 0;
}

```