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:
import java.util.*;
import java.io.*;


public class Main{


  public static void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    Integer[] arr = Arrays.stream(in.readLine().split(" ")).map(Integer::parseInt).toArray(Integer[]::new);

    int n = arr[0], m = arr[1];
    Integer[][] mat = new Integer[n][m];

    for(int i=0;i<n;i++){
      arr = Arrays.stream(in.readLine().split(" ")).map(Integer::parseInt).toArray(Integer[]::new);
      mat[i] = arr;
    }
    List<String> result = new ArrayList<>();
    if(n<m){
      for(int i=0;i<n;i++){
        int min=500;
        for(int j=0;j<m;j++){
          if(min>mat[i][j]){
            min=mat[i][j];
          }
        }
        for(int j=0;j<min;j++){
          result.add("row "+(i+1));
        }
        for(int j=0;j<m;j++){
          mat[i][j]-=min;
        }
      }
      for(int i=0;i<m;i++){
        int min=500;
        for(int j=0;j<n;j++){
          if(min>mat[j][i]){
            min=mat[j][i];
          }
        }
        for(int j=0;j<min;j++){
          result.add("col "+(i+1));
        }
        for(int j=0;j<n;j++){
          mat[j][i]-=min;
        }
      }
    }else{
      for(int i=0;i<m;i++){
        int min=500;
        for(int j=0;j<n;j++){
          if(min>mat[j][i]){
            min=mat[j][i];
          }
        }
        for(int j=0;j<min;j++){
          result.add("col "+(i+1));
        }
        for(int j=0;j<n;j++){
          mat[j][i]-=min;
        }
      }
      for(int i=0;i<n;i++){
        int min=500;
        for(int j=0;j<m;j++){
          if(min>mat[i][j]){
            min=mat[i][j];
          }
        }
        for(int j=0;j<min;j++){
          result.add("row "+(i+1));
        }
        for(int j=0;j<m;j++){
          mat[i][j]-=min;
        }
      }
    }

    int max=0;
    for(int i=0;i<m;i++){
      for(int j=0;j<n;j++){
        if(max<mat[j][i]){
          max=mat[j][i];
        }
      }
    }
    if(max>0){
      System.out.println(-1);
      return;
    }
    System.out.println(result.size());
    for (String s: result ) {
      System.out.println(s);
    }
  }
}