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 JAVA):

Solution:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskC {
  static int[][] mat;
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();

    String[] s = br.readLine().split(" ");
    int n  = Integer.parseInt(s[0]);
    int m  = Integer.parseInt(s[1]);

    mat = new int[n][m];
    for (int i=0; i<n; i++) {
      s = br.readLine().split(" ");
      int sum = 0;
      for (int j=0; j<m; j++) {
        mat[i][j] = Integer.parseInt(s[j]);
        sum += mat[i][j];
      }
    }
    
    int[] rVals = new int[n];
    int[] cVals = new int[m];
    int bestSum = 987654321;
    int[] bestR = new int[n];
    int[] bestC = new int[m];
    for (int i=0; i<=500; i++) {
      Arrays.fill(rVals,-1);
      Arrays.fill(cVals,-1);
      rVals[0] = i;
      boolean valid = true;
      int sum = 0;
      for (int j=0; j<m; j++) {
        cVals[j] = mat[0][j]-i;
        if (cVals[j]<0) {
          valid = false;
        }
        sum += cVals[j];
      }
      for (int j=0; j<n; j++) {
        rVals[j] = mat[j][0]-cVals[0];
        if (rVals[j]<0)
          valid = false;
        sum += rVals[j];
      }
      if (!valid)
        continue;
      for (int j=0; j<n; j++) {
        if (!valid)
          break;
        for (int k=0; k<m; k++) {
          if (rVals[j]+cVals[k]!=mat[j][k]) {
            valid = false;
            break;
          }
        }
      }
      if (!valid)
        continue;
      if (sum<bestSum) {
        bestR = Arrays.copyOf(rVals, n);
        bestC = Arrays.copyOf(cVals, m);
        bestSum = sum;
      }
    }
    if (bestSum==987654321) {
      System.out.println("-1");
      System.exit(0);
    }
    System.out.println(bestSum);
    for (int i=0; i<n; i++) {
      int val = bestR[i];
      for (int j=0; j<val; j++) {
        sb.append("row "+(i+1));
        sb.append("\n");
      }
    }
    for (int i=0; i<m; i++) {
      int val = bestC[i];
      for (int j=0; j<val; j++) {
        sb.append("col "+(i+1));
        sb.append("\n");
      }
    }
    System.out.print(sb);

  }
}