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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Hashtable;


public class problem1 {
	
	public static void main(String[]args) throws IOException {
		BufferedReader l = new BufferedReader(new InputStreamReader(System.in));
		String[] t = (l.readLine()).split(" ");
		int n = Integer.parseInt(t[0]);int m = Integer.parseInt(t[1]);
		boolean[][] dat = new boolean[n][m];
		boolean[][] use = new boolean[n][m];
		int[][] out = new int[n][m];
		boolean[] nn = new boolean[n];
		boolean[] mm = new boolean[m];
		for(int i=0;i<n;i++) {
			t = (l.readLine()).split(" ");
			for(int j=0;j<m;j++) {
				if(t[j].equals("0"))
					dat[i][j] = false;
				else dat[i][j] = true;
				use[i][j] = true;
				out[i][j] = 1;
				mm[j] = true;
			}
			nn[i] = true;
		}
		
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++) {
				if(!dat[i][j]){
					if(mm[j]){
						for(int k=0;k<n;k++){
							use[k][j] = false;
							out[k][j] = 0;
						}
						mm[j] = false;
					}
					if(nn[i]){
						for(int k=0;k<m;k++){
							use[i][k] = false;
							out[i][k] = 0;
						}
						nn[i] = false;
					}
				}
			}
		}
		for(int i=0;i<n;i++) {
			for(int j=0;j<m;j++) {
				if(dat[i][j]){
					boolean found = false;
					if(mm[j]){
						for(int k=0;k<n;k++){
							if(use[k][j]){
								use[k][j] = false;
								out[k][j] = 1;
								found = true;
								break;
							}
							else if(out[k][j] == 1)
								found = true;
						}
					}
					else if(nn[i]){
						for(int k=0;k<m;k++){
							if(use[i][k]){
								use[i][k] = false;
								out[i][k] = 1;
								found = true;
								break;
							}
							else if(out[i][k] == 1)
								found = true;
						}
					}
					else {
						System.out.println("NO");
						return;
					}
					if(!found){
						System.out.println("NO");
						return;
					}
				}
			}
		}
		System.out.println("YES");
		for(int i =0;i<n;i++){
			for(int j=0;j<m-1;j++){
				System.out.print(out[i][j]+" ");
			}
			System.out.println(out[i][m-1]);
		}
	}	
		
}