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 Div2_419_C {
	static boolean [] zeroRow, zeroCol;
	public static void update(int i, int j, int [][] grid) {
		grid[i][j]--;
		if (grid[i][j] == 0) {
			zeroRow[i] = true;
			zeroCol[j] = true;
		}
	}
	public static void getRows(int [][] grid, ArrayList<String> ans) {
		for (int i = 0;i < grid.length;i++) {
			while(!zeroRow[i]) {
				ans.add("row "+(i+1));
				for (int j = 0;j < grid[i].length;j++) {
					update(i, j, grid);
				}
			}
		}
	}
	public static void getCols(int [][] grid, ArrayList<String> ans) {
		for (int j = 0;j < grid[0].length;j++) {
			while(!zeroCol[j]) {
				ans.add("col "+(j + 1));
				for (int i = 0;i < grid.length;i++) {
					update(i, j, grid);
				}
			}
		}
	}
	public static void main(String[] args) throws IOException {
		BufferedInputStream bis = new BufferedInputStream(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(bis));
		PrintWriter out = new PrintWriter(System.out);
		String line;
		StringTokenizer st;
		while ((line = br.readLine()) != null && !line.equals("")) {
			st = new StringTokenizer(line);
			int n = Integer.parseInt(st.nextToken());
			int m = Integer.parseInt(st.nextToken());
			int [][] grid = new int[n][m];
			zeroRow = new boolean[n];
			zeroCol = new boolean[m];
			for (int i = 0;i < grid.length;i++) {
				st = new StringTokenizer(br.readLine());
				for (int j = 0; j < grid[i].length; j++) {
					grid[i][j] = Integer.parseInt(st.nextToken());
					if (grid[i][j] == 0) {
						zeroRow[i] = true;
						zeroCol[j] = true;
					}
				}
			}
			ArrayList<String> ans = new ArrayList<>();
			if (n < m) {
				getRows(grid, ans);
				getCols(grid, ans);
			} else {
				getCols(grid, ans);
				getRows(grid, ans);
			}
			boolean hasAns = true;
			for (int i = 0;i < grid.length && hasAns;i++) {
				for (int j = 0;j < grid[i].length && hasAns;j++) {
					if (grid[i][j] != 0)
						hasAns = false;
				}
			}
			if (hasAns) {
				out.println(ans.size());
				for (String a : ans)
					out.println(a);
			} else {
				out.println(-1);
			}
		}
		out.close();
	}
}