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.InputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

/**
 */
public class Main816C {
	public static void main(String[] args) {
		run(System.in, System.out);
	}

	public static void run(InputStream in, PrintStream out) {

		try (Scanner sc = new Scanner(in)) {
			int n = sc.nextInt();
			int m = sc.nextInt();

			int[][] tab = new int[n][m];
			int[] row = new int[n];
			int[] col = new int[m];
			for (int i = 0; i < n; i++)
				row[i] = Integer.MAX_VALUE;
			for (int i = 0; i < m; i++)
				col[i] = Integer.MAX_VALUE;

			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					int k = sc.nextInt();
					tab[i][j] = k;
					if (k < row[i])
						row[i] = k;
					if (k < col[j])
						col[j] = k;
				}
			}

			System.err.println(Arrays.toString(row));
			System.err.println(Arrays.toString(col));
			int cpt = 0;
			StringBuilder res = new StringBuilder();

			if (n < m) {
				for (int i = 0; i < n; i++) {
					cpt += row[i];
					for (int k = 0; k < row[i]; k++) {
						res.append("row ").append(i + 1).append('\n');
					}
				}
				for (int j = 0; j < m; j++) {
					for (int i = 1; i < n; i++) {
						if (tab[i][j] - row[i] != tab[0][j] - row[0]) {
							out.print("-1");
							return;
						}
					}
					cpt += tab[0][j] - row[0];
					for (int k = 0; k < tab[0][j] - row[0]; k++) {
						res.append("col ").append(j + 1).append('\n');
					}
				}
			} else {
				for (int j = 0; j < m; j++) {
					cpt += col[j];
					for (int k = 0; k < col[j]; k++) {
						res.append("col ").append(j + 1).append('\n');
					}
				}
				for (int i = 0; i < n; i++) {
					for (int j = 1; j < m; j++) {
						if (tab[i][j] - col[j] != tab[i][0] - col[0]) {
							out.print("-1");
							return;
						}
					}
					cpt += tab[i][0] - col[0];
					for (int k = 0; k < tab[i][0] - col[0]; k++) {
						res.append("row ").append(i + 1).append('\n');
					}
				}
			}

			out.print(cpt + "\n" + res.toString());
		}
	}

}