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} 
#include <bits/stdc++.h>
using namespace std;
int a[107][107], b[107][107];
inline void fill1(int i, int j) {
  b[i][j] = b[i + 1][j] = b[i][j + 1] = b[i + 1][j + 1] = 1;
  return;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (auto i = (1); i <= (n); ++i) {
    for (auto j = (1); j <= (m); ++j) scanf("%d", &a[i][j]);
  }
  vector<pair<int, int> > ans;
  for (auto i = (1); i <= (n); ++i) {
    for (auto j = (1); j <= (m); ++j) {
      if (a[i][j] && !b[i][j]) {
        if (a[i + 1][j] && a[i][j + 1] && a[i + 1][j + 1])
          ans.emplace_back(i, j), fill1(i, j);
        else if (a[i - 1][j] && a[i - 1][j + 1] && a[i][j + 1])
          ans.emplace_back(i - 1, j), fill1(i - 1, j);
        else if (a[i - 1][j - 1] && a[i - 1][j] && a[i][j - 1])
          ans.emplace_back(i - 1, j - 1), fill1(i - 1, j - 1);
        else
          ans.emplace_back(i, j - 1), fill1(i, j - 1);
      }
    }
  }
  for (auto i = (1); i <= (n); ++i) {
    for (auto j = (1); j <= (m); ++j) {
      if (a[i][j] != b[i][j]) {
        puts("-1");
        return 0;
      }
    }
  }
  for (auto i = (0); i <= ((int)ans.size() - 1); ++i) {
    if (ans[i].first < 1 || ans[i].first >= n || ans[i].second < 1 ||
        ans[i].second >= m) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", (int)ans.size());
  for (auto i = (0); i <= ((int)ans.size() - 1); ++i)
    printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}
