#include <bits/stdc++.h>
using namespace std;
const int mx = 1e2 + 9;
const int inf = 1e18 + 9;
int A[mx][mx], B[mx][mx], freq[mx], fr[mx];
vector<pair<int, int> > vp1, vp2;
vector<int> v1, v2;
bool sign[mx], sign2[mx];
map<int, int> mp;
string s1, s2, s3;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  bool f = false;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> A[i][j];
    }
  }
  bool f1 = false;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      if (A[i][j] == 0 || A[i][j + 1] == 0 || A[i + 1][j] == 0 ||
          A[i + 1][j + 1] == 0)
        f1 = true;
      if (!f1) {
        B[i][j] = 1;
        B[i + 1][j] = 1;
        B[i][j + 1] = 1;
        B[i + 1][j + 1] = 1;
        vp1.push_back({i, j});
      }
      f1 = false;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (A[i][j] != B[i][j]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << vp1.size() << endl;
  for (int i = 0; i < vp1.size(); i++)
    cout << vp1[i].first << " " << vp1[i].second << endl;
}

### Prompt

Please formulate a CPP solution 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;
const int mx = 1e2 + 9;
const int inf = 1e18 + 9;
int A[mx][mx], B[mx][mx], freq[mx], fr[mx];
vector<pair<int, int> > vp1, vp2;
vector<int> v1, v2;
bool sign[mx], sign2[mx];
map<int, int> mp;
string s1, s2, s3;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  bool f = false;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> A[i][j];
    }
  }
  bool f1 = false;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      if (A[i][j] == 0 || A[i][j + 1] == 0 || A[i + 1][j] == 0 ||
          A[i + 1][j + 1] == 0)
        f1 = true;
      if (!f1) {
        B[i][j] = 1;
        B[i + 1][j] = 1;
        B[i][j + 1] = 1;
        B[i + 1][j + 1] = 1;
        vp1.push_back({i, j});
      }
      f1 = false;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (A[i][j] != B[i][j]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << vp1.size() << endl;
  for (int i = 0; i < vp1.size(); i++)
    cout << vp1[i].first << " " << vp1[i].second << endl;
}

```