#include <bits/stdc++.h>
using namespace std;
int a[51][51];
pair<int, int> mem[51][51];
bool isValid(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
bool isQuad(pair<int, int> w, pair<int, int> x, pair<int, int> y,
            pair<int, int> z, int n, int m) {
  return isValid(w.first, w.second, n, m) && isValid(x.first, x.second, n, m) &&
         isValid(y.first, y.second, n, m) && isValid(z.first, z.second, n, m) &&
         a[w.first][w.second] && a[x.first][x.second] && a[y.first][y.second] &&
         a[z.first][z.second];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, tot = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j]) {
        tot++;
        if (isQuad({i, j}, {i + 1, j}, {i + 1, j + 1}, {i, j + 1}, n, m)) {
          mem[i][j] = {i + 1, j + 1};
        } else if (isQuad({i, j}, {i, j - 1}, {i + 1, j}, {i + 1, j - 1}, n,
                          m)) {
          mem[i][j] = {i + 1, j};
        } else if (isQuad({i, j}, {i - 1, j}, {i - 1, j + 1}, {i, j + 1}, n,
                          m)) {
          mem[i][j] = {i, j + 1};
        } else if (isQuad({i, j}, {i, j - 1}, {i - 1, j}, {i - 1, j - 1}, n,
                          m)) {
          mem[i][j] = {i, j};
        } else {
          cout << -1 << "\n";
          return 0;
        }
      }
    }
  }
  cout << tot << "\n";
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j]) cout << mem[i][j].first << " " << mem[i][j].second << "\n";
}

### Prompt

Please create a solution in Cpp 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 a[51][51];
pair<int, int> mem[51][51];
bool isValid(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
bool isQuad(pair<int, int> w, pair<int, int> x, pair<int, int> y,
            pair<int, int> z, int n, int m) {
  return isValid(w.first, w.second, n, m) && isValid(x.first, x.second, n, m) &&
         isValid(y.first, y.second, n, m) && isValid(z.first, z.second, n, m) &&
         a[w.first][w.second] && a[x.first][x.second] && a[y.first][y.second] &&
         a[z.first][z.second];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, tot = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j]) {
        tot++;
        if (isQuad({i, j}, {i + 1, j}, {i + 1, j + 1}, {i, j + 1}, n, m)) {
          mem[i][j] = {i + 1, j + 1};
        } else if (isQuad({i, j}, {i, j - 1}, {i + 1, j}, {i + 1, j - 1}, n,
                          m)) {
          mem[i][j] = {i + 1, j};
        } else if (isQuad({i, j}, {i - 1, j}, {i - 1, j + 1}, {i, j + 1}, n,
                          m)) {
          mem[i][j] = {i, j + 1};
        } else if (isQuad({i, j}, {i, j - 1}, {i - 1, j}, {i - 1, j - 1}, n,
                          m)) {
          mem[i][j] = {i, j};
        } else {
          cout << -1 << "\n";
          return 0;
        }
      }
    }
  }
  cout << tot << "\n";
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j]) cout << mem[i][j].first << " " << mem[i][j].second << "\n";
}

```