#include <bits/stdc++.h>
using namespace std;
int mp[105][105];
int mp1[105][105];
int n, m;
int num[2][500000];
int main() {
  int ans = 0;
  int ans1 = 0;
  cin >> n >> m;
  int cnt = 0;
  int cnt1 = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &mp[i][j]), mp1[i][j] = mp[i][j];
  for (int i = 1; i <= n; i++) {
    int Min = 1e9;
    for (int j = 1; j <= m; j++) {
      Min = min(Min, mp[i][j]);
    }
    if (Min) {
      ans += Min;
      for (int j = 1; j <= Min; j++) {
        num[0][cnt++] = i;
      }
      for (int j = 1; j <= m; j++) mp[i][j] -= Min;
    }
  }
  for (int i = 1; i <= m; i++) {
    int Min = 1e9;
    for (int j = 1; j <= n; j++) {
      Min = min(Min, mp[j][i]);
    }
    if (Min) {
      ans += Min;
      for (int j = 1; j <= Min; j++) {
        num[0][cnt++] = -i;
      }
      for (int j = 1; j <= n; j++) mp[j][i] -= Min;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (mp[i][j] > 0) {
        return printf("-1\n"), 0;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    int Min = 1e9;
    for (int j = 1; j <= n; j++) {
      Min = min(Min, mp1[j][i]);
    }
    if (Min) {
      ans1 += Min;
      for (int j = 1; j <= Min; j++) {
        num[1][cnt1++] = -i;
      }
      for (int j = 1; j <= n; j++) mp1[j][i] -= Min;
    }
  }
  for (int i = 1; i <= n; i++) {
    int Min = 1e9;
    for (int j = 1; j <= m; j++) {
      Min = min(Min, mp1[i][j]);
    }
    if (Min) {
      ans1 += Min;
      for (int j = 1; j <= Min; j++) {
        num[1][cnt1++] = i;
      }
      for (int j = 1; j <= m; j++) mp1[i][j] -= Min;
    }
  }
  int id;
  int xx;
  cout << min(ans, ans1) << endl;
  if (ans < ans1)
    id = 0, xx = cnt;
  else
    id = 1, xx = cnt1;
  for (int i = 0; i < xx; i++) {
    if (num[id][i] > 0) {
      printf("row %d\n", num[id][i]);
    } else {
      printf("col %d\n", -num[id][i]);
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int mp[105][105];
int mp1[105][105];
int n, m;
int num[2][500000];
int main() {
  int ans = 0;
  int ans1 = 0;
  cin >> n >> m;
  int cnt = 0;
  int cnt1 = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &mp[i][j]), mp1[i][j] = mp[i][j];
  for (int i = 1; i <= n; i++) {
    int Min = 1e9;
    for (int j = 1; j <= m; j++) {
      Min = min(Min, mp[i][j]);
    }
    if (Min) {
      ans += Min;
      for (int j = 1; j <= Min; j++) {
        num[0][cnt++] = i;
      }
      for (int j = 1; j <= m; j++) mp[i][j] -= Min;
    }
  }
  for (int i = 1; i <= m; i++) {
    int Min = 1e9;
    for (int j = 1; j <= n; j++) {
      Min = min(Min, mp[j][i]);
    }
    if (Min) {
      ans += Min;
      for (int j = 1; j <= Min; j++) {
        num[0][cnt++] = -i;
      }
      for (int j = 1; j <= n; j++) mp[j][i] -= Min;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (mp[i][j] > 0) {
        return printf("-1\n"), 0;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    int Min = 1e9;
    for (int j = 1; j <= n; j++) {
      Min = min(Min, mp1[j][i]);
    }
    if (Min) {
      ans1 += Min;
      for (int j = 1; j <= Min; j++) {
        num[1][cnt1++] = -i;
      }
      for (int j = 1; j <= n; j++) mp1[j][i] -= Min;
    }
  }
  for (int i = 1; i <= n; i++) {
    int Min = 1e9;
    for (int j = 1; j <= m; j++) {
      Min = min(Min, mp1[i][j]);
    }
    if (Min) {
      ans1 += Min;
      for (int j = 1; j <= Min; j++) {
        num[1][cnt1++] = i;
      }
      for (int j = 1; j <= m; j++) mp1[i][j] -= Min;
    }
  }
  int id;
  int xx;
  cout << min(ans, ans1) << endl;
  if (ans < ans1)
    id = 0, xx = cnt;
  else
    id = 1, xx = cnt1;
  for (int i = 0; i < xx; i++) {
    if (num[id][i] > 0) {
      printf("row %d\n", num[id][i]);
    } else {
      printf("col %d\n", -num[id][i]);
    }
  }
  return 0;
}

```