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 Dfs {

    public static void main(String[] args)throws Exception {
     //Scanner in =new Scanner(System.in);
     BufferedReader in =new BufferedReader(new InputStreamReader(System.in));
     StringTokenizer tok = new StringTokenizer(in.readLine());
     int m =Integer.parseInt(tok.nextToken());int n=Integer.parseInt(tok.nextToken());
     
     int arr[][] =new int [m][n];
    
    int one=0;
     for(int i=0;i<m;i++){
         tok = new StringTokenizer(in.readLine());
         for(int j =0;j<n;j++){
             arr[i][j]=Integer.parseInt(tok.nextToken());if(arr[i][j]==1)one++;} }
     int cnt =0;int p =0;int l =0;
   
     
     int r =0; int c =0; StringBuilder str = new StringBuilder();
     for(int i= 0;i<m;i++){
         for(int j= 0;j<n;j++){
             
             for(int k=0;k<m;k++)
                 if(arr[k][j]==arr[i][j])
                     c++;
                 else 
                     cnt++;
                 for(int k=0;k<n;k++)
                     if(arr[i][k]==arr[i][j])
                         r++;
                     else p++;
       
                 if(p!=0||cnt!=0)
                 if(cnt>0&&p>0&&arr[i][j]==1){
          
                     System.out.println("NO");System.exit(0);
                 }
                 else if(arr[i][j]==0&&cnt==m-1&&p==n-1){
                    
                     System.out.println("NO");System.exit(0);
                 }
                
                 if((r!=n||c!=m)&&arr[i][j]==1)
                     l++;
              
                 if(c==m&&r==n&&arr[i][j]==1){
                     str.append(1+" ");}
                 else if (c==m&&r==n&&arr[i][j]==0){
                    str.append(0+" ");}
                 else { str.append(0+" ");}
                 c=0;r=0;cnt= 0;p =0;
                
}str.append("\n");
}  
        
   if(l==one&&one!=0){
        System.out.println("NO");System.exit(0);
    
}
 
        System.out.println("YES");
        System.out.println(str);

     
          }
    static double nextDouble(StringTokenizer tok){
    return Double.parseDouble(tok.nextToken());
}
public static int nextInt( StringTokenizer s) throws Exception {
        return Integer.parseInt(s.nextToken());
    }
public static long nextLong(StringTokenizer bu) throws Exception {
        return Long.parseLong(bu.nextToken());
    }}