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:

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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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, i, j, k, a[505][505];
  int mini, ans, ans2, minii[505], req[505], flag, flag2, flag3, minimum_pos,
      needed;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  for (i = 0; i < m; i++) {
    mini = a[0][i];
    for (j = 0; j < n; j++) {
      mini = min(mini, a[j][i]);
    }
    minii[i] = mini;
  }
  ans = 1e9;
  for (i = 0; i <= 500; i++) {
    ans2 = i;
    flag = 0;
    flag3 = 0;
    for (j = 0; j < m; j++) {
      if (i > a[0][j]) {
        flag = 1;
        break;
      }
      req[j] = a[0][j] - i;
      if (req[j] > minii[j]) {
        flag3 = 1;
        break;
      }
      ans2 += req[j];
    }
    if (flag) {
      break;
    }
    if (flag3) {
      continue;
    }
    flag2 = 0;
    for (j = 1; j < n; j++) {
      for (k = 0; k < m; k++) {
        if (k == 0) {
          if (a[j][k] < req[k]) {
            flag2 = 1;
            break;
          } else {
            needed = a[j][k] - req[k];
          }
        } else {
          if (needed != (a[j][k] - req[k])) {
            flag2 = 1;
            break;
          }
        }
      }
      if (flag2) {
        break;
      } else {
        ans2 += needed;
      }
    }
    if (flag2) {
      continue;
    }
    if (ans > ans2) {
      ans = ans2;
      minimum_pos = i;
    }
  }
  if (ans == 1e9) {
    cout << -1;
  } else {
    cout << ans << "\n";
    for (i = 0; i < minimum_pos; i++) {
      cout << "row " << 1 << "\n";
    }
    for (j = 0; j < m; j++) {
      req[j] = a[0][j] - minimum_pos;
      for (k = 0; k < a[0][j] - minimum_pos; k++) {
        cout << "col " << j + 1 << "\n";
      }
    }
    for (i = 1; i < n; i++) {
      for (j = 0; j < a[i][0] - req[0]; j++) {
        cout << "row " << i + 1 << "\n";
      }
    }
  }
}