Problem: Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:

<image> where <image> is equal to 1 if some ai = 1, otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

<image>.

(Bij is OR of all elements in row i and column j of matrix A)

Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100), number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print m rows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Examples

Input

2 2
1 0
0 0


Output

NO


Input

2 3
1 1 1
1 1 1


Output

YES
1 1 1
1 1 1


Input

2 3
0 1 0
1 1 1


Output

YES
0 0 0
0 1 0

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class B {
	static StringBuilder st = new StringBuilder();
	
		
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt() , m  = sc.nextInt();
				
		boolean [][] A = new boolean [n][m];
		boolean [][] B = new boolean [n][m];
		
		for(int i = 0 ; i < n ; i++)
			Arrays.fill(A[i], true);
		
		for(int i = 0  ; i < n ; i++ )
			for(int j = 0  ; j < m ; j++ )
			{
				B[i][j] = sc.nextInt() == 1 ; 
				
				if(!B[i][j])
				{
					for(int k = 0; k < m ; k++)
						A[i][k] = false ; 
					
					for(int k = 0; k < n ; k++)
						A[k][j] = false ; 
					
				}
			}
		
		
		boolean [][] res = new boolean [n][m];
		
		for(int i = 0 ; i < n ;i++)
			for(int j = 0 ; j < m ; j++)
			{
				for(int k = 0; k < m ; k++)
					res[i][j] |= A[i][k];
				
				for(int k = 0; k < n ; k++)
					res[i][j] |= A[k][j];
				
			}
		
		boolean can = true ; 
		for(int i = 0 ; i < n ;i++)
			for(int j = 0 ; j < m ; j++)
				can &= B[i][j] == res[i][j];
		
		
		if(!can)
			out.println("NO");
		else 
		{
			out.println("YES");
			
			for(int i = 0 ; i < n ;i++)
			{
				for(int j = 0 ; j < m ; j++)
					out.print(A[i][j] ? "1 " : "0 ");
				
				out.println();
			}
			
		}
 		
		
			
		out.flush();
		out.close();

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		public Scanner(String path) throws Exception {
			br = new BufferedReader(new FileReader(path));
		}

		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 double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

	}

	static void shuffle(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			int tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}

}