#include <bits/stdc++.h>
using namespace std;
const int maxn = 51;
int n, m;
int a[51][51];
int v[51][51];
struct p {
  int x, y;
} c[3000];
bool ok(int x, int y) {
  if (a[x][y] && a[x + 1][y] && a[x][y + 1] && a[x + 1][y + 1]) return 1;
  return 0;
}
void st(int x, int y) {
  v[x][y] = v[x + 1][y] = v[x][y + 1] = v[x + 1][y + 1] = 1;
}
void print() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) printf("%d ", v[i][j]);
    printf("\n");
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  int ct = 0;
  for (int i = 1; i < n; i++)
    for (int j = 1; j < m; j++) {
      if (a[i][j] == 0) continue;
      if (v[i][j]) {
        if (ok(i, j)) {
          ct++;
          c[ct].x = i, c[ct].y = j;
          st(i, j);
        }
      } else {
        if (ok(i, j)) {
          ct++;
          c[ct].x = i, c[ct].y = j;
          st(i, j);
        } else {
          printf("-1");
          return 0;
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i][m] == 1 && v[i][m] == 0) {
      printf("-1");
      return 0;
    }
  for (int i = 1; i <= m; i++)
    if (a[n][i] == 1 && v[n][i] == 0) {
      printf("-1");
      return 0;
    }
  printf("%d\n", ct);
  for (int i = 1; i <= ct; i++) printf("%d %d\n", c[i].x, c[i].y);
  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 int maxn = 51;
int n, m;
int a[51][51];
int v[51][51];
struct p {
  int x, y;
} c[3000];
bool ok(int x, int y) {
  if (a[x][y] && a[x + 1][y] && a[x][y + 1] && a[x + 1][y + 1]) return 1;
  return 0;
}
void st(int x, int y) {
  v[x][y] = v[x + 1][y] = v[x][y + 1] = v[x + 1][y + 1] = 1;
}
void print() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) printf("%d ", v[i][j]);
    printf("\n");
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  int ct = 0;
  for (int i = 1; i < n; i++)
    for (int j = 1; j < m; j++) {
      if (a[i][j] == 0) continue;
      if (v[i][j]) {
        if (ok(i, j)) {
          ct++;
          c[ct].x = i, c[ct].y = j;
          st(i, j);
        }
      } else {
        if (ok(i, j)) {
          ct++;
          c[ct].x = i, c[ct].y = j;
          st(i, j);
        } else {
          printf("-1");
          return 0;
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i][m] == 1 && v[i][m] == 0) {
      printf("-1");
      return 0;
    }
  for (int i = 1; i <= m; i++)
    if (a[n][i] == 1 && v[n][i] == 0) {
      printf("-1");
      return 0;
    }
  printf("%d\n", ct);
  for (int i = 1; i <= ct; i++) printf("%d %d\n", c[i].x, c[i].y);
  return 0;
}

```