#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int k = n * m;
  long long int arr[k + 1];
  arr[0] = -1;
  map<long long int, long long int> mp1;
  long long int t = 0, sum = 0;
  for (long long int i = 1; i <= k; i++) {
    cin >> arr[i];
    if (arr[i] == 1)
      mp1[i] = 8;
    else
      t++;
  }
  if (t == k) {
    cout << "0";
    return 0;
  }
  long long int p = k - m, x, c = 0;
  vector<long long int> v;
  map<long long int, long long int> mp;
  for (long long int i = 1; i < p; i++) {
    if (arr[i] == 1 && i % m != 0) {
      x = m * ((i / m) + 1) + (i % m);
      if (arr[i + 1] == 1 && arr[x] == 1 && arr[x + 1] == 1) {
        mp[i] = 8;
        mp[i + 1] = 8;
        mp[x] = 8;
        mp[x + 1] = 8;
        sum += 4;
        v.push_back(i);
        c++;
      }
    }
  }
  sum = mp.size();
  long long int t1 = mp1.size();
  if (sum != t1)
    cout << "-1";
  else {
    cout << c << "\n";
    for (long long int i = 0; i < v.size(); i++) {
      cout << (v[i] / m) + 1 << " " << (v[i] % m) << "\n";
    }
  }
}

### Prompt

Develop a solution in cpp 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>
using namespace std;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int k = n * m;
  long long int arr[k + 1];
  arr[0] = -1;
  map<long long int, long long int> mp1;
  long long int t = 0, sum = 0;
  for (long long int i = 1; i <= k; i++) {
    cin >> arr[i];
    if (arr[i] == 1)
      mp1[i] = 8;
    else
      t++;
  }
  if (t == k) {
    cout << "0";
    return 0;
  }
  long long int p = k - m, x, c = 0;
  vector<long long int> v;
  map<long long int, long long int> mp;
  for (long long int i = 1; i < p; i++) {
    if (arr[i] == 1 && i % m != 0) {
      x = m * ((i / m) + 1) + (i % m);
      if (arr[i + 1] == 1 && arr[x] == 1 && arr[x + 1] == 1) {
        mp[i] = 8;
        mp[i + 1] = 8;
        mp[x] = 8;
        mp[x + 1] = 8;
        sum += 4;
        v.push_back(i);
        c++;
      }
    }
  }
  sum = mp.size();
  long long int t1 = mp1.size();
  if (sum != t1)
    cout << "-1";
  else {
    cout << c << "\n";
    for (long long int i = 0; i < v.size(); i++) {
      cout << (v[i] / m) + 1 << " " << (v[i] % m) << "\n";
    }
  }
}

```