#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const long long MAX = 1e5 + 7;
const long long dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const long long dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
long long i, j, k, t, n, q, l, r, mid;
bool flag;
long long m;
bool a[51][51];
set<pair<long long, long long> > v;
void solve() {
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) cin >> a[i][j];
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      flag = 0;
      if (a[i][j]) {
        if (i + 1 < n && j + 1 < m && a[i][j + 1] && a[i + 1][j] &&
            a[i + 1][j + 1]) {
          flag = 1;
          v.insert(make_pair(i, j));
        }
        if (i - 1 >= 0 && j + 1 < m && a[i - 1][j] && a[i - 1][j + 1] &&
            a[i][j + 1]) {
          flag = 1;
          v.insert(make_pair(i - 1, j));
        }
        if (i + 1 < n && j - 1 >= 0 && a[i + 1][j - 1] && a[i + 1][j] &&
            a[i][j - 1]) {
          flag = 1;
          v.insert(make_pair(i, j - 1));
        }
        if (i - 1 >= 0 && j - 1 >= 0 && a[i - 1][j] && a[i - 1][j - 1] &&
            a[i][j - 1]) {
          flag = 1;
          v.insert(make_pair(i - 1, j - 1));
        }
        if (flag == 0) {
          cout << "-1" << endl;
          return;
        }
      }
    }
  }
  cout << v.size() << endl;
  for (auto x : v) {
    cout << x.first + 1 << " " << x.second + 1 << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  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;
const long long MOD = 998244353;
const long long MAX = 1e5 + 7;
const long long dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const long long dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
long long i, j, k, t, n, q, l, r, mid;
bool flag;
long long m;
bool a[51][51];
set<pair<long long, long long> > v;
void solve() {
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) cin >> a[i][j];
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      flag = 0;
      if (a[i][j]) {
        if (i + 1 < n && j + 1 < m && a[i][j + 1] && a[i + 1][j] &&
            a[i + 1][j + 1]) {
          flag = 1;
          v.insert(make_pair(i, j));
        }
        if (i - 1 >= 0 && j + 1 < m && a[i - 1][j] && a[i - 1][j + 1] &&
            a[i][j + 1]) {
          flag = 1;
          v.insert(make_pair(i - 1, j));
        }
        if (i + 1 < n && j - 1 >= 0 && a[i + 1][j - 1] && a[i + 1][j] &&
            a[i][j - 1]) {
          flag = 1;
          v.insert(make_pair(i, j - 1));
        }
        if (i - 1 >= 0 && j - 1 >= 0 && a[i - 1][j] && a[i - 1][j - 1] &&
            a[i][j - 1]) {
          flag = 1;
          v.insert(make_pair(i - 1, j - 1));
        }
        if (flag == 0) {
          cout << "-1" << endl;
          return;
        }
      }
    }
  }
  cout << v.size() << endl;
  for (auto x : v) {
    cout << x.first + 1 << " " << x.second + 1 << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}

```