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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class R419C
{
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[][] g = new int[n][m];
		for (int i = 0; i < g.length; i++)
			for (int j = 0; j < g[i].length; j++)
				g[i][j] = sc.nextInt();
		
		int cnt = 0;
		StringBuilder sb = new StringBuilder("");
		if(n < m)
		{
			nxt: for (int i = 0; i < n; i++)
			{
				while(true)
				{
					boolean zero = false;
					for (int j = 0; j < m; j++)
						if(g[i][j] == 0)
						{
							zero = true;
							break;
						}
					if(zero)
						continue nxt;
					
					sb.append("row " + (i+1) + "\n");
					cnt++;
					for (int j = 0; j < m; j++)
						g[i][j]--;
				}
			}
			
			nxt: for (int i = 0; i < m; i++)
			{
				while(true)
				{
					boolean zero = false;
					for (int j = 0; j < n; j++)
						if(g[j][i] == 0)
						{
							zero = true;
							break;
						}
					if(zero)
						continue nxt;
					
					sb.append("col " + (i+1) + "\n");
					cnt++;
					for (int j = 0; j < n; j++)
						g[j][i]--;
				}
			}
		}
		else
		{
			nxt: for (int i = 0; i < m; i++)
			{
				while(true)
				{
					boolean zero = false;
					for (int j = 0; j < n; j++)
						if(g[j][i] == 0)
						{
							zero = true;
							break;
						}
					if(zero)
						continue nxt;
					
					sb.append("col " + (i+1) + "\n");
					cnt++;
					for (int j = 0; j < n; j++)
						g[j][i]--;
				}
			}
			
			nxt: for (int i = 0; i < n; i++)
			{
				while(true)
				{
					boolean zero = false;
					for (int j = 0; j < m; j++)
						if(g[i][j] == 0)
						{
							zero = true;
							break;
						}
					if(zero)
						continue nxt;
					
					sb.append("row " + (i+1) + "\n");
					cnt++;
					for (int j = 0; j < m; j++)
						g[i][j]--;
				}
			}
		}
		
		boolean valid = true;
		outer: for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
			{
				if(g[i][j] != 0)
				{
					valid = false;
					break outer;
				}
			}
		if(!valid)
			pw.println(-1);
		else
		{
			pw.println(cnt);
			pw.print(sb);
		}
		pw.flush();
		pw.close();
	}
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}
		
		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}
		
		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}
		
		public boolean ready() throws IOException {return br.ready();}
	}
	
	
	static class Pair implements Comparable<Pair>
	{
		int x, y;
		public Pair(int x, int y)
		{
			this.x = x;
			this.y = y;
		}
		
		public int compareTo(Pair p)
		{
			return 0;
		}
	}
}