#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[50][50], b[50][50] = {0};
vector<pair<int, int> > op;
int c[4][2] = {{-1, -1}, {-1, +1}, {+1, -1}, {+1, +1}};
bool is_needed(int x, int y) {
  bool is_needed = false;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      is_needed = is_needed || (a[x + i][y + j] == 1 && b[x + i][y + j] == 0);
  return is_needed;
}
bool is_valid(int i, int j) { return (i >= 0 && i < n && j >= 0 && j < m); }
bool maybe_fit(int x, int y) {
  int l, r;
  bool valid;
  l = 1;
  r = 1;
  valid = true;
  for (int j = 0; j < 2; j++) {
    for (int k = 0; k < 2; k++) {
      valid = valid && is_valid(x + l * j, y + r * k) &&
              (a[x + l * j][y + r * k] == 1);
    }
  }
  if (valid) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++) b[x + l * j][y + r * k] = 1;
    return true;
  }
  return false;
}
int main() {
  int i, j;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < m - 1; j++) {
      if (is_needed(i, j) && maybe_fit(i, j)) {
        op.push_back(make_pair(i, j));
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] != b[i][j]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << op.size() << endl;
  for (i = 0; i < op.size(); i++) {
    cout << op[i].first + 1 << " " << op[i].second + 1 << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, m;
int a[50][50], b[50][50] = {0};
vector<pair<int, int> > op;
int c[4][2] = {{-1, -1}, {-1, +1}, {+1, -1}, {+1, +1}};
bool is_needed(int x, int y) {
  bool is_needed = false;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      is_needed = is_needed || (a[x + i][y + j] == 1 && b[x + i][y + j] == 0);
  return is_needed;
}
bool is_valid(int i, int j) { return (i >= 0 && i < n && j >= 0 && j < m); }
bool maybe_fit(int x, int y) {
  int l, r;
  bool valid;
  l = 1;
  r = 1;
  valid = true;
  for (int j = 0; j < 2; j++) {
    for (int k = 0; k < 2; k++) {
      valid = valid && is_valid(x + l * j, y + r * k) &&
              (a[x + l * j][y + r * k] == 1);
    }
  }
  if (valid) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++) b[x + l * j][y + r * k] = 1;
    return true;
  }
  return false;
}
int main() {
  int i, j;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < m - 1; j++) {
      if (is_needed(i, j) && maybe_fit(i, j)) {
        op.push_back(make_pair(i, j));
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] != b[i][j]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << op.size() << endl;
  for (i = 0; i < op.size(); i++) {
    cout << op[i].first + 1 << " " << op[i].second + 1 << endl;
  }
  return 0;
}

```