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.
import java.util.Scanner;

public class Main {

    public static void main(String[] args) throws Exception {

        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt(),m=sc.nextInt();
        int[][] a = new int[101][101];
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                a[i][j] = sc.nextInt();
            }
        }

        int[] cr=null,lr=null;
        int min = Integer.MAX_VALUE;

        for(int i=0;i<=a[1][1];i++){
            int[] c=new int[n+1];
            int[] l=new int[m+1];
            boolean right = true;

            c[1] = i;
            l[1] = a[1][1] - c[1];
            if(l[1] < 0)
                continue;
            int sum = c[1]+l[1];

            for(int j=2;j<=n;j++){
                c[j] = a[j][1]-a[1][1]+c[1];
                if(c[j] < 0){
                    right = false;
                    break;
                }
                sum+=c[j];
            }
            for(int j=2;j<=m;j++){
                l[j] = a[1][j]-a[1][1]+l[1];
                if(l[j] < 0){
                    right = false;
                    break;
                }
                sum+=l[j];
            }

            if(right){
                for(int t=2;t<=n;t++){
                    for(int s=2;s<=m;s++){
                        if(c[t]+l[s] != a[t][s]){
                            right = false;
                            break;
                        }
                    }
                }
            }

            if(right && sum < min){
                min = sum;
                cr = c;
                lr = l;
            }
        }

        if(min == Integer.MAX_VALUE){
            System.out.println(-1);
        }else{
            System.out.println(min);
            for(int i=1;i<=n;i++){
                for(int j=0;j<cr[i];j++){
                    System.out.println("row "+i);
                }
            }
            for(int i=1;i<=m;i++){
                for(int j=0;j<lr[i];j++){
                    System.out.println("col "+i);
                }
            }
        }
    }
}
