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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        private int cnt;
        private List<String> tempOut;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            cnt = 0;
            tempOut = new ArrayList<>();
            int n = in.nextInt();
            int m = in.nextInt();
            int[][] g = new int[n][];
            for (int i = 0; i < n; i++) {
                g[i] = new int[m];
                for (int j = 0; j < m; j++) {
                    g[i][j] = in.nextInt();
                }
            }

            if (n < m) {
                doRow(g, n, m);
                doCol(g, n, m);
            } else {
                doCol(g, n, m);
                doRow(g, n, m);
            }

            if (done(g, n, m)) {
                out.write(cnt + "\n");
                for (String line : tempOut) {
                    out.write(line);
                    out.write("\n");
                }
            } else {
                out.write("-1");
                out.write("\n");
            }

        }

        private static boolean done(int[][] g, int n, int m) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (g[i][j] != 0) {
                        return false;
                    }
                }
            }
            return true;
        }

        private void doCol(int[][] g, int n, int m) {
            for (int j = 0; j < m; j++) {
                int min = 5000;
                for (int i = 0; i < n; i++) {
                    min = Math.min(min, g[i][j]);
                }
                if (min > 0) {
                    for (int k = 0; k < n; k++) {
                        g[k][j] -= min;
                    }
                }
                for (int k = 0; k < min; k++) {
                    cnt++;
                    tempOut.add("col " + (j + 1));
                }
            }
        }

        private void doRow(int[][] g, int n, int m) {
            for (int i = 0; i < n; i++) {
                int min = 5000;
                for (int j = 0; j < m; j++) {
                    min = Math.min(min, g[i][j]);
                }
                if (min > 0) {
                    for (int k = 0; k < m; k++) {
                        g[i][k] -= min;
                    }
                }
                for (int k = 0; k < min; k++) {
                    cnt++;
                    tempOut.add("row " + (i + 1));
                }
            }
        }

    }
}