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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  long long i, j, k;
  long long arr[105][105];
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> arr[i][j];
    }
  }
  long long ansrr[105], anscc[105];
  long long ansr[105], ansc[105];
  long long count = INT_MAX;
  for (ansr[1] = 0; ansr[1] <= 500; ansr[1]++) {
    long long prec = 0;
    prec += ansr[1];
    bool flag = true;
    long long pre = ansr[1];
    for (j = 1; j <= m; j++) {
      ansc[j] = arr[1][j] - pre;
      if (ansc[j] < 0)
        flag = false;
      else
        prec += ansc[j];
    }
    pre = ansc[1];
    for (i = 2; i <= n; i++) {
      ansr[i] = arr[i][1] - pre;
      if (ansr[i] < 0)
        flag = false;
      else
        prec += ansr[i];
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        if (arr[i][j] != ansr[i] + ansc[j]) flag = false;
      }
    }
    if (flag && count > prec) {
      count = prec;
      for (i = 1; i <= n; i++) {
        ansrr[i] = ansr[i];
      }
      for (i = 1; i <= m; i++) {
        anscc[i] = ansc[i];
      }
    }
  }
  if (count == INT_MAX) {
    cout << "-1";
    return 0;
  }
  cout << count << "\n";
  for (i = 1; i <= n; i++) {
    while (ansrr[i]--) {
      cout << "row " << i << "\n";
    }
  }
  for (i = 1; i <= m; i++) {
    while (anscc[i]--) {
      cout << "col " << i << "\n";
    }
  }
  return 0;
}
