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.*;


public class Main {
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int m = scanner.nextInt();
		int[][] a = new int[n][m];
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < m; j++) {
				a[i][j] = scanner.nextInt();
			}
		}	
		int[] rows = new int[n];
		int[] columns = new int[m];
		int count = 0;
		if(m > n) {
			for(int i = 0; i < n; i++) {
				rows[i] = getMin(a, i, -1);
				count += rows[i];
			}
			for(int i = 0; i < m; i++) {
				columns[i] = getMin(a, -1, i);
				count += columns[i];
			}
		} else {
			for(int i = 0; i < m; i++) {
				columns[i] = getMin(a, -1, i);
				count += columns[i];
			}
			for(int i = 0; i < n; i++) {
				rows[i] = getMin(a, i, -1);
				count += rows[i];
			}
		}
		if(isEmpty(a)) {
			System.out.println(count);
			for(int i = 0; i < n; i++) {
				for(int j = 0; j < rows[i]; j++) {
					System.out.println("row " + (i+1));
				}

			}
			for(int i = 0; i < m; i++) {
				for(int j = 0; j < columns[i]; j++) {
					System.out.println("col " + (i+1));
				}

			}
		} else {
			System.out.println(-1);
		}
		
	}
	
	
	private static boolean isEmpty(int[][] a) {
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a[0].length; j++) {
				if(a[i][j] != 0) return false;
			}
		}
		return true;
	}


	static int getMin(int[][] a, int row, int column) {
		int min = Integer.MAX_VALUE;
		if(row == -1) {
			for(int j = 0; j < a.length; j++) {
				min = Math.min(min, a[j][column]);
			}
			for(int j = 0; j < a.length; j++) {
				a[j][column] -= min;
			}
		} else {
			for(int j = 0; j < a[0].length; j++) {
				min = Math.min(min, a[row][j]);
			}
			for(int j = 0; j < a[0].length; j++) {
				a[row][j] -= min;
			}
		}
		return min;
	}
}