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;
int arr[110][110];
int row1[10010];
int row2[10010];
int col[10010];
int col2[10010];
int n, m;
bool check(int i, int j) {
  if (arr[i][j] == (row1[i] + col2[j]))
    return 1;
  else
    return 0;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) scanf("%d", &arr[i][j]);
  }
  int ans1 = 0;
  int ans2 = 0;
  int maxrow = -1000000;
  for (int i = 0; i < n; ++i) {
    int minn = 100000;
    for (int j = 0; j < m; ++j) minn = min(arr[i][j], minn);
    row1[i] = minn;
    ans1 += minn;
    maxrow = max(minn, maxrow);
  }
  for (int i = 0; i < m; ++i) {
    int maxx = -100000;
    for (int j = 0; j < n; ++j) maxx = max(arr[j][i], maxx);
    ans1 += max(maxx - maxrow, 0);
    col2[i] = max(maxx - maxrow, 0);
  }
  int maxcol = -1000000;
  for (int i = 0; i < m; ++i) {
    int minn = 100000;
    for (int j = 0; j < n; ++j) minn = min(arr[j][i], minn);
    col[i] = minn;
    ans2 += minn;
    maxcol = max(minn, maxcol);
  }
  for (int i = 0; i < n; ++i) {
    int maxx = -100000;
    for (int j = 0; j < m; ++j) maxx = max(arr[i][j], maxx);
    ans2 += max(maxx - maxcol, 0);
    row2[i] = max(maxx - maxcol, 0);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j)
      if (!check(i, j)) return puts("-1");
  }
  if (ans1 <= ans2) {
    cout << ans1 << endl;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < row1[i]; ++j) cout << "row " << i + 1 << endl;
    }
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < col2[i]; ++j) cout << "col " << i + 1 << endl;
    }
  } else {
    cout << ans2 << endl;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < row2[i]; ++j) cout << "row " << i + 1 << endl;
    }
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < col[i]; ++j) cout << "col " << i + 1 << endl;
    }
  }
  return 0;
}
