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

public class C {
    public static void main(String[] args) throws Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int rs = Integer.parseInt(st.nextToken());
        int cs = Integer.parseInt(st.nextToken());
        
        int[][] res = new int[rs][cs];
        for (int i = 0; i < rs; i++)
        {
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < cs; j++)
            {
                res[i][j] = Integer.parseInt(st.nextToken());
            }
        }
        ArrayList<Integer> rowComs = new ArrayList<>();
        ArrayList<Integer> colComs = new ArrayList<>();
        
        boolean ch = true;
        while (ch)
        {
            ch = false;
            for (int c = 0; c < cs - 1; c++)
            {
                while (res[0][c] > res[0][c + 1])
                {
                    colComs.add(c);
                    for (int i = 0; i < rs; i++)
                    {
                        res[i][c]--;
                    }
                    ch = true;
                }
                while (res[0][c] < res[0][c + 1])
                {
                    colComs.add(c + 1);
                    for (int i = 0; i < rs; i++)
                    {
                        res[i][c + 1]--;
                    }
                    ch = true;
                }
            }
        }
        
        // at this point all rows should be equal
        
        // if all above 1, remove whichever side is smallest
        int min = 500;
        for (int i = 0; i < rs; i++)
        {
            if (res[i][0] < min) min = res[i][0];
        }
        if (min != 0)
        {
            if (rs < cs)
            {
                for (int i = 0; i < rs; i++)
                {
                    for (int j = 0; j < min; j++)
                    {
                        rowComs.add(i);
                    }
                    for (int j = 0; j < cs; j++)
                    {
                        res[i][j] -= min;
                    }
                }
            }
            else
            {
                for (int j = 0; j < cs; j++)
                {
                    for (int i = 0; i < min; i++)
                    {
                        colComs.add(j);
                    }
                    for (int i = 0; i < rs; i++)
                    {
                        res[i][j] -= min;
                    }
                }
            }
        }
        
        // remove rows
        for (int i = 0; i < rs; i++)
        {
            int t = res[i][0];
            for (int k = 0; k < t; k++)
            {
                rowComs.add(i);
                for (int j = 0; j < cs; j++)
                {
                    res[i][j]--;
                }
            }
        }
        
        // check if valid
        boolean valid = true;
        out: for (int i = 0; i < rs; i++)
        {
            for (int j = 0; j < cs; j++)
            {
                if (res[i][j] != 0)
                {
                    valid = false;
                    break out;
                }
            }
        }
        
        if (valid)
        {
            System.out.println(rowComs.size() + colComs.size());
            for (int i : rowComs)
            {
                System.out.println("row " + (i + 1));
            }
            for (int j : colComs)
            {
                System.out.println("col " + (j + 1));
            }
        }
        else
        {
            System.out.println("-1");
        }
    }
}