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.util.Arrays;
import java.util.Scanner;


public class OrinTheMatrix {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner input = new Scanner(System.in);
		int m = input.nextInt();
		int n = input.nextInt();
		char[][] s = new char[m][n];
		for(int i=0; i<m;i++){
	       for(int j=0; j<n;j++)s[i][j]= input.next().charAt(0);
			
		}
		for(int i=0; i<m;i++){
			for(int j=0; j<n;j++){
				if(s[i][j]=='1' && !checkRow(s,i) && !checkColumn(s,j)){
					System.out.println("NO");
					return;
				}
			}
		}
		//System.out.println("fdf");
		char[][] tmp = new char[m][n];
		for(int i=0; i<m;i++)Arrays.fill(tmp[i],'0');
		for(int i=0; i<m;i++){
			if(s[i][0]=='1'){
				if(checkRow(s,i)){
					boolean condition = false;
					for(int j=0; j<n;j++){
						if(checkColumn(s,j)){
							condition = true;
						   	tmp[i][j] = '1';
						}
					}
					if(!condition){
						System.out.println("NO");
						return;
					}
				}
				
			}
		}
		for(int j=0; j<n;j++){
			if(checkColumn(s,j)){
				boolean condition = false;
				for(int i=0; i<m;i++){
					if(checkRow(s,i)){
						condition = true;
						tmp[i][j] = '1';
					}
				}
				if(!condition){
					System.out.println("NO");
					return;
				}
			}
		}
		System.out.println("YES");
		for(int i=0; i<m;i++){
			for(int j=0; j<n;j++){
				System.out.print(tmp[i][j]+" ");
			}
			System.out.println();
		}
	//	System.out.println("fd");
		
		
	}

	static boolean checkRow(char[][] s, int i){
		for(int j=0; j<s[i].length;j++){
			if(s[i][j]!='1')return false;
		}
		return true;
	}
	static boolean checkColumn(char[][]s, int j){
		for(int i=0; i<s.length;i++){
			if(s[i][j]!='1')return false;
		}
		return true;
	}
}