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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 111;
int n, m;
int a[maxn][maxn], row[maxn], col[maxn], b[maxn][maxn], row2[maxn], col2[maxn];
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]);
      b[i][j] = a[i][j];
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int flag = 0;
    while (1) {
      for (int j = 1; j <= m; j++) {
        if (a[i][j] == 0) flag = 1;
      }
      if (flag) break;
      for (int j = 1; j <= m; j++) {
        a[i][j]--;
      }
      cnt++;
      row[i]++;
    }
  }
  for (int i = 1; i <= m; i++) {
    int flag = 0;
    while (1) {
      for (int j = 1; j <= n; j++) {
        if (a[j][i] == 0) flag = 1;
      }
      if (flag) break;
      for (int j = 1; j <= n; j++) {
        a[j][i]--;
      }
      cnt++;
      col[i]++;
    }
  }
  int cnt2 = 0;
  for (int i = 1; i <= m; i++) {
    int flag = 0;
    while (1) {
      for (int j = 1; j <= n; j++) {
        if (b[j][i] == 0) flag = 1;
      }
      if (flag) break;
      for (int j = 1; j <= n; j++) {
        b[j][i]--;
      }
      cnt2++;
      col2[i]++;
    }
    for (int i = 1; i <= n; i++) {
      int flag = 0;
      while (1) {
        for (int j = 1; j <= m; j++) {
          if (b[i][j] == 0) flag = 1;
        }
        if (flag) break;
        for (int j = 1; j <= m; j++) {
          b[i][j]--;
        }
        cnt2++;
        row2[i]++;
      }
    }
  }
  int ok = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j] != 0) ok = 0;
    }
  }
  if (!ok)
    printf("-1\n");
  else if (cnt < cnt2) {
    printf("%d\n", cnt);
    for (int i = 1; i <= n; i++) {
      while (row[i]) {
        printf("row %d\n", i);
        row[i]--;
      }
    }
    for (int i = 1; i <= m; i++) {
      while (col[i]) {
        printf("col %d\n", i);
        col[i]--;
      }
    }
  } else {
    printf("%d\n", cnt2);
    for (int i = 1; i <= n; i++) {
      while (row2[i]) {
        printf("row %d\n", i);
        row2[i]--;
      }
    }
    for (int i = 1; i <= m; i++) {
      while (col2[i]) {
        printf("col %d\n", i);
        col2[i]--;
      }
    }
  }
  return 0;
}