#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  setIO();
  int n, m;
  cin >> n >> m;
  int g[n][m];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> g[i][j];
    }
  }
  int ng[n][m];
  memset(ng, 0, sizeof(ng));
  vector<pii> ans;
  for (int i = 0; i < n - 1; ++i) {
    for (int j = 0; j < m - 1; ++j) {
      if (g[i][j] == 1 && g[i + 1][j] == 1 && g[i][j + 1] == 1 &&
          g[i + 1][j + 1] == 1) {
        ng[i][j] = 1;
        ng[i + 1][j] = 1;
        ng[i][j + 1] = 1;
        ng[i + 1][j + 1] = 1;
        ans.emplace_back(make_pair(i, j));
      }
    }
  }
  bool ok = true;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (ng[i][j] != g[i][j]) ok = false;
    }
  }
  if (!ok) {
    cout << "-1\n";
  } else {
    cout << ans.size() << "\n";
    for (auto i : ans) {
      cout << i.first + 1 << " " << i.second + 1 << "\n";
    }
  }
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  setIO();
  int n, m;
  cin >> n >> m;
  int g[n][m];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> g[i][j];
    }
  }
  int ng[n][m];
  memset(ng, 0, sizeof(ng));
  vector<pii> ans;
  for (int i = 0; i < n - 1; ++i) {
    for (int j = 0; j < m - 1; ++j) {
      if (g[i][j] == 1 && g[i + 1][j] == 1 && g[i][j + 1] == 1 &&
          g[i + 1][j + 1] == 1) {
        ng[i][j] = 1;
        ng[i + 1][j] = 1;
        ng[i][j + 1] = 1;
        ng[i + 1][j + 1] = 1;
        ans.emplace_back(make_pair(i, j));
      }
    }
  }
  bool ok = true;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (ng[i][j] != g[i][j]) ok = false;
    }
  }
  if (!ok) {
    cout << "-1\n";
  } else {
    cout << ans.size() << "\n";
    for (auto i : ans) {
      cout << i.first + 1 << " " << i.second + 1 << "\n";
    }
  }
}

```