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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 *
 * @author Don Li
 */
public class KarenGame3 {
    
    int N = 110;
    
    int n, m;
    int[][] g;
    
    int[] row = new int[N], col = new int[N];
    
    void solve() {
        n = in.nextInt();
        m = in.nextInt();
        g = new int[n][m];
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) g[i][j] = in.nextInt();
        
        int[] p = findEmptyCell();
        if (p != null && findPlan(p[0], p[1])) {
            printPlan();
            return;
        }
        
        if (n <= m) {
            int min = g[0][0];
            for (int j = 1; j < m; j++) min = Math.min(min, g[0][j]);
            int y = -1;
            for (int j = 0; j < m; j++) {
                g[0][j] -= min;
                if (g[0][j] == 0) y = j;
            }
            if (findPlan(0, y)) {
                row[0] += min;
                printPlan();
            } else {
                out.println(-1);
            }
        } else {
            int min = g[0][0];
            for (int i = 1; i < n; i++) min = Math.min(min, g[i][0]);
            int x = -1;
            for (int i = 0; i < n; i++) {
                g[i][0] -= min;
                if (g[i][0] == 0) x = i;
            }
            if (findPlan(x, 0)) {
                col[0] += min;
                printPlan();
            } else {
                out.println(-1);
            }
        }
    }
    
    int[] findEmptyCell() {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (g[i][j] == 0) {
                    return new int[]{i, j};
                }
            }
        }
        return null;
    }
    
    boolean findPlan(int x, int y) {
        for (int i = 0; i < n; i++) {
            if (i != x) row[i] = g[i][y];
        }
        for (int j = 0; j < m; j++) {
            if (j != y) col[j] = g[x][j];
        }
        
        for (int i = 0; i < n; i++) if (row[i] < 0) return false;
        for (int j = 0; j < m; j++) if (col[j] < 0) return false;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (g[i][j] != row[i] + col[j]) {
                    return false;
                }
            }
        }
        return true;
    }
    
    void printPlan() {
        int tot = 0;
        for (int i = 0; i < n; i++) tot += row[i];
        for (int j = 0; j < m; j++) tot += col[j];
        
        out.println(tot);
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < row[i]; j++) {
                out.printf("row %d%n", i + 1);
            }
        }
        for (int j = 0; j < m; j++) {
            for (int i = 0; i < col[j]; i++) {
                out.printf("col %d%n", j + 1);
            }
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new KarenGame3().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}