#include <bits/stdc++.h>
const int N = 100005;
const long double pi = 3.141592653;
const long long MOD = 1000000007;
using namespace std;
int dx[9] = {1, -1, 0, 0, 1, 1, -1, -1, 0};
int dy[9] = {0, 0, 1, -1, 1, -1, 1, -1, 0};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    int n, m, i, j, mat[55][55];
    cin >> n >> m;
    vector<pair<int, int> > v;
    for (i = 0; i < 55; i++) {
      for (j = 0; j < 55; j++) mat[i][j] = 0;
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) cin >> mat[i][j];
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        if (mat[i][j] == 1) {
          if (mat[i + 1][j] == 1 && mat[i][j + 1] == 1 &&
              mat[i + 1][j + 1] == 1) {
            if (i < n && j < m) v.push_back({i, j});
          } else if (mat[i + 1][j] == 1 && mat[i][j - 1] == 1 &&
                     mat[i + 1][j - 1] == 1)
            continue;
          else if (mat[i][j - 1] == 1 && mat[i - 1][j] == 1 &&
                   mat[i - 1][j - 1] == 1)
            continue;
          else if (mat[i][j + 1] == 1 && mat[i - 1][j] == 1 &&
                   mat[i - 1][j + 1] == 1)
            continue;
          else
            break;
        }
      }
      if (j <= m) break;
    }
    if (i <= n)
      cout << -1;
    else {
      cout << (long long)v.size() << "\n";
      for (i = 0; i < (long long)v.size(); i++)
        cout << v[i].first << " " << v[i].second << "\n";
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
const int N = 100005;
const long double pi = 3.141592653;
const long long MOD = 1000000007;
using namespace std;
int dx[9] = {1, -1, 0, 0, 1, 1, -1, -1, 0};
int dy[9] = {0, 0, 1, -1, 1, -1, 1, -1, 0};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    int n, m, i, j, mat[55][55];
    cin >> n >> m;
    vector<pair<int, int> > v;
    for (i = 0; i < 55; i++) {
      for (j = 0; j < 55; j++) mat[i][j] = 0;
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) cin >> mat[i][j];
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        if (mat[i][j] == 1) {
          if (mat[i + 1][j] == 1 && mat[i][j + 1] == 1 &&
              mat[i + 1][j + 1] == 1) {
            if (i < n && j < m) v.push_back({i, j});
          } else if (mat[i + 1][j] == 1 && mat[i][j - 1] == 1 &&
                     mat[i + 1][j - 1] == 1)
            continue;
          else if (mat[i][j - 1] == 1 && mat[i - 1][j] == 1 &&
                   mat[i - 1][j - 1] == 1)
            continue;
          else if (mat[i][j + 1] == 1 && mat[i - 1][j] == 1 &&
                   mat[i - 1][j + 1] == 1)
            continue;
          else
            break;
        }
      }
      if (j <= m) break;
    }
    if (i <= n)
      cout << -1;
    else {
      cout << (long long)v.size() << "\n";
      for (i = 0; i < (long long)v.size(); i++)
        cout << v[i].first << " " << v[i].second << "\n";
    }
  }
  return 0;
}

```