#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 100;
const int md = 1000000007;
priority_queue<int, vector<int>, greater<int> > pq;
bool compare(pair<int, int> &a, pair<int, int> &b) { return a.first > b.first; }
int main() {
  int n, m, arr[60][60], ans[60][60] = {};
  bool sama = true;
  vector<pair<int, int> > path;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i += 1)
    for (int j = 0; j < m; j += 1) scanf("%d", &arr[i][j]);
  for (int i = 0; i < n - 1; i += 1) {
    for (int j = 0; j < m - 1; j += 1) {
      if (arr[i][j] + arr[i + 1][j] + arr[i][j + 1] + arr[i + 1][j + 1] == 4) {
        path.push_back(make_pair(i, j));
        ans[i][j] = ans[i + 1][j] = ans[i][j + 1] = ans[i + 1][j + 1] = 1;
      }
    }
  }
  for (int i = 0; i < n; i += 1) {
    for (int j = 0; j < m; j += 1)
      if (arr[i][j] != ans[i][j]) sama = false;
  }
  if (sama) {
    printf("%d\n", path.size());
    for (int i = 0; i < path.size(); i += 1)
      printf("%d %d\n", path[i].first + 1, path[i].second + 1);
  } else
    printf("-1\n");
}

### 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;
const int mx = 2e5 + 100;
const int md = 1000000007;
priority_queue<int, vector<int>, greater<int> > pq;
bool compare(pair<int, int> &a, pair<int, int> &b) { return a.first > b.first; }
int main() {
  int n, m, arr[60][60], ans[60][60] = {};
  bool sama = true;
  vector<pair<int, int> > path;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i += 1)
    for (int j = 0; j < m; j += 1) scanf("%d", &arr[i][j]);
  for (int i = 0; i < n - 1; i += 1) {
    for (int j = 0; j < m - 1; j += 1) {
      if (arr[i][j] + arr[i + 1][j] + arr[i][j + 1] + arr[i + 1][j + 1] == 4) {
        path.push_back(make_pair(i, j));
        ans[i][j] = ans[i + 1][j] = ans[i][j + 1] = ans[i + 1][j + 1] = 1;
      }
    }
  }
  for (int i = 0; i < n; i += 1) {
    for (int j = 0; j < m; j += 1)
      if (arr[i][j] != ans[i][j]) sama = false;
  }
  if (sama) {
    printf("%d\n", path.size());
    for (int i = 0; i < path.size(); i += 1)
      printf("%d %d\n", path[i].first + 1, path[i].second + 1);
  } else
    printf("-1\n");
}

```