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:

<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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
 
public class TestClass {
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args) throws Exception{
        
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s[] = in.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int m = Integer.parseInt(s[1]);
        int r[] = new int[n];
        int a[][] = new int[n][m];
        for(int i=0;i<n;i++)
        {
        	String s1[] = in.readLine().split(" ");
        	for(int j=0;j<m;j++)
        	{
        		a[i][j] = Integer.parseInt(s1[j]);
        	}
        }
        if(n<=m)
        {
        ArrayList b = new ArrayList();
        ArrayList c = new ArrayList();
        for(int i=0;i<n;i++)
        {
        	int min = Integer.MAX_VALUE;
        	for(int j=0;j<m;j++)
        	{
        		min = Math.min(min, a[i][j]);
        	}
        	if(min>0)
        	{
        		for(int j=0;j<m;j++)
            	{
            		a[i][j]-=min;
            	}
        		while(min>0)
        		{
	        		b.add(i+1);
	        		c.add("row");
	        		min--;
        		}
        	}
        }
        for(int j=0;j<m;j++)
        {
        	int min = Integer.MAX_VALUE;
        	for(int i=0;i<n;i++)
        	{
        		min = Math.min(min, a[i][j]);
        	}
        	if(min>0)
        	{
        		for(int i=0;i<n;i++)
            	{
            		a[i][j]-=min;
            	}
        		while(min>0)
        		{
        		b.add(j+1);
        		c.add("col");
        		min--;
        		}
        	}

        }
        for(int i=0;i<n;i++)
        {
        	for(int j=0;j<m;j++)
        	{
        	if(a[i][j]!=0)
        	{
        		out.println(-1);
        		out.close();
        		System.exit(0);
        	}
        	}
        }
        out.println(b.size());
        for(int i=0;i<b.size();i++)
        {
        	out.println(c.get(i)+" "+b.get(i));
        }
        }
        else
        {
        	  ArrayList b = new ArrayList();
              ArrayList c = new ArrayList();
              for(int j=0;j<m;j++)
              {
              	int min = Integer.MAX_VALUE;
              	for(int i=0;i<n;i++)
              	{
              		min = Math.min(min, a[i][j]);
              	}
              	if(min>0)
              	{
              		for(int i=0;i<n;i++)
                  	{
                  		a[i][j]-=min;
                  	}
              		while(min>0)
              		{
              		b.add(j+1);
              		c.add("col");
              		min--;
              		}
              	}

              }
              
              for(int i=0;i<n;i++)
              {
              	int min = Integer.MAX_VALUE;
              	for(int j=0;j<m;j++)
              	{
              		min = Math.min(min, a[i][j]);
              	}
              	if(min>0)
              	{
              		for(int j=0;j<m;j++)
                  	{
                  		a[i][j]-=min;
                  	}
              		while(min>0)
              		{
      	        		b.add(i+1);
      	        		c.add("row");
      	        		min--;
              		}
              	}
              }
              for(int i=0;i<n;i++)
              {
              	for(int j=0;j<m;j++)
              	{
              	if(a[i][j]!=0)
              	{
              		out.println(-1);
              		out.close();
              		System.exit(0);
              	}
              	}
              }
              out.println(b.size());
              for(int i=0;i<b.size();i++)
              {
              	out.println(c.get(i)+" "+b.get(i));
              }
        }
        out.close();        
    }
   
 }