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 read() {
  char ch = getchar();
  int w = 1, c = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) c = (c << 1) + (c << 3) + (ch ^ 48);
  return w * c;
}
int n, m;
int a[100][100];
bool op[100][100];
bool check(int x, int y, int dx, int dy) {
  if (a[x + dx][y + dy] && a[x][y + dy] && a[x + dx][y]) {
    return 1;
  }
  return 0;
}
int main() {
  n = read();
  m = read();
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (m); ++j) a[i][j] = read();
  memset(op, 0, sizeof op);
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (m); ++j) {
      if (a[i][j]) {
        bool fl = 0;
        if (check(i, j, -1, -1)) {
          op[i - 1][j - 1] = 1;
          fl = 1;
        } else if (check(i, j, 1, -1)) {
          op[i][j - 1] = 1;
          fl = 1;
        } else if (check(i, j, -1, 1)) {
          op[i - 1][j] = 1;
          fl = 1;
        } else if (check(i, j, 1, 1)) {
          op[i][j] = 1;
          fl = 1;
        }
        if (!fl) {
          puts("-1");
          return 0;
        }
      }
    }
  }
  vector<pair<int, int> > ans;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (m); ++j) {
      if (op[i][j]) {
        ans.push_back(make_pair(i, j));
      }
    }
  cout << ((int)ans.size() - 1) + 1 << "\n";
  for (int i = (0); i <= (((int)ans.size() - 1)); ++i) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}
