#include <bits/stdc++.h>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, n, m, k, cnt = 0, ans = 0, t = 1;
  cin >> n >> m;
  long long a[n][m];
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) cin >> a[i][j];
  long long b[n][m];
  memset(b, 0, sizeof(b));
  vector<pair<long long, long long> > out;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (i + 1 < n && j + 1 < m) {
        if (a[i][j] == 1 && a[i][j + 1] == 1 && a[i + 1][j] == 1 &&
            a[i + 1][j + 1] == 1) {
          b[i][j] = 1;
          b[i + 1][j] = 1;
          b[i][j + 1] = 1;
          b[i + 1][j + 1] = 1;
          out.push_back(make_pair(i, j));
        }
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] != b[i][j]) return cout << -1, 0;
    }
  }
  cout << out.size() << '\n';
  for (auto it : out) cout << it.first + 1 << " " << it.second + 1 << '\n';
  return 0;
}

### 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>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, n, m, k, cnt = 0, ans = 0, t = 1;
  cin >> n >> m;
  long long a[n][m];
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) cin >> a[i][j];
  long long b[n][m];
  memset(b, 0, sizeof(b));
  vector<pair<long long, long long> > out;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (i + 1 < n && j + 1 < m) {
        if (a[i][j] == 1 && a[i][j + 1] == 1 && a[i + 1][j] == 1 &&
            a[i + 1][j + 1] == 1) {
          b[i][j] = 1;
          b[i + 1][j] = 1;
          b[i][j + 1] = 1;
          b[i + 1][j + 1] = 1;
          out.push_back(make_pair(i, j));
        }
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] != b[i][j]) return cout << -1, 0;
    }
  }
  cout << out.size() << '\n';
  for (auto it : out) cout << it.first + 1 << " " << it.second + 1 << '\n';
  return 0;
}

```