#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int n, m, a[N][N], vis[N][N], p[N * N], q[N * N], top;
bool check(int x, int y) {
  if (!(a[x][y] && a[x][y + 1] && a[x + 1][y] && a[x + 1][y + 1])) return 0;
  vis[x][y] = vis[x][y + 1] = vis[x + 1][y] = vis[x + 1][y + 1] = 1;
  p[++top] = x, q[top] = y;
  return 1;
}
bool find(int x, int y) {
  if (y > 1 && x < n && check(x, y - 1)) return 1;
  if (y < m && x < n && check(x, y)) return 1;
  if (y > 1 && x > 1 && check(x - 1, y - 1)) return 1;
  if (y < m && x > 1 && check(x - 1, y)) return 1;
  return 0;
}
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]);
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (m); j++)
      if (a[i][j]) {
        if (vis[i][j]) continue;
        if (!find(i, j)) return puts("-1"), 0;
      }
  printf("%d\n", top);
  for (int i = (1); i <= (top); i++) printf("%d %d\n", p[i], q[i]);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
using namespace std;
const int N = 55;
int n, m, a[N][N], vis[N][N], p[N * N], q[N * N], top;
bool check(int x, int y) {
  if (!(a[x][y] && a[x][y + 1] && a[x + 1][y] && a[x + 1][y + 1])) return 0;
  vis[x][y] = vis[x][y + 1] = vis[x + 1][y] = vis[x + 1][y + 1] = 1;
  p[++top] = x, q[top] = y;
  return 1;
}
bool find(int x, int y) {
  if (y > 1 && x < n && check(x, y - 1)) return 1;
  if (y < m && x < n && check(x, y)) return 1;
  if (y > 1 && x > 1 && check(x - 1, y - 1)) return 1;
  if (y < m && x > 1 && check(x - 1, y)) return 1;
  return 0;
}
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]);
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (m); j++)
      if (a[i][j]) {
        if (vis[i][j]) continue;
        if (!find(i, j)) return puts("-1"), 0;
      }
  printf("%d\n", top);
  for (int i = (1); i <= (top); i++) printf("%d %d\n", p[i], q[i]);
  return 0;
}

```