#include <bits/stdc++.h>
using namespace std;
template <class code>
inline code read(const code &a) {
  code x = 0;
  short w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -x : x;
}
const int N = 55;
const int dx[4] = {0, 1, 0, 1};
const int dy[4] = {0, 0, 1, 1};
bool check[N][N];
bool change[N][N];
struct Point {
  int x, y;
} f[N * N];
int len;
int main() {
  int n = read(n);
  int m = read(m);
  memset(check, 1, sizeof(check));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) check[i][j] = read(check[i][j]);
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      bool ok = 1;
      for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
        ok &= check[x][y];
      }
      if (ok) {
        for (int k = 0; k < 4; k++) {
          int x = i + dx[k];
          int y = j + dy[k];
          change[x][y] = 1;
        }
        f[++len].x = i;
        f[len].y = j;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++)
      if (change[i][j] != check[i][j]) {
        puts("-1");
        return 0;
      }
  }
  printf("%d\n", len);
  for (int i = 1; i <= len; i++) printf("%d %d\n", f[i].x, f[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;
template <class code>
inline code read(const code &a) {
  code x = 0;
  short w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -x : x;
}
const int N = 55;
const int dx[4] = {0, 1, 0, 1};
const int dy[4] = {0, 0, 1, 1};
bool check[N][N];
bool change[N][N];
struct Point {
  int x, y;
} f[N * N];
int len;
int main() {
  int n = read(n);
  int m = read(m);
  memset(check, 1, sizeof(check));
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) check[i][j] = read(check[i][j]);
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      bool ok = 1;
      for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
        ok &= check[x][y];
      }
      if (ok) {
        for (int k = 0; k < 4; k++) {
          int x = i + dx[k];
          int y = j + dy[k];
          change[x][y] = 1;
        }
        f[++len].x = i;
        f[len].y = j;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++)
      if (change[i][j] != check[i][j]) {
        puts("-1");
        return 0;
      }
  }
  printf("%d\n", len);
  for (int i = 1; i <= len; i++) printf("%d %d\n", f[i].x, f[i].y);
  return 0;
}

```