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 b277{
   public static void main(String[] args) {
      MyScanner obj = new MyScanner();
      int m=obj.nextInt();
      int n=obj.nextInt();
      int c[][]=new int[m][n];
      int a[][]=new int[m][n];
      int b[][]=new int[m][n];
      for(int i=0;i<m;i++)
      {
		  Arrays.fill(a[i],-1);
		  Arrays.fill(c[i],0);
	  }
      for(int i=0;i<m;i++)
      {
		  for(int j=0;j<n;j++)
		  {
		       b[i][j]=obj.nextInt();
			   if(b[i][j]==0)
			   {
				   for(int k=0;k<n;k++)
				   a[i][k]=0;
				   for(int k=0;k<m;k++)
				   a[k][j]=0;
			    }
			}
	  }
	  for(int i=0;i<m;i++)
	  {
		  for(int j=0;j<n;j++)
		  {
			  if(a[i][j]!=0)
			  a[i][j]=1;
		  }
	  }
	  int flag=0;
	  for(int i=0;i<m;i++)
	  {
		  for(int j=0;j<n;j++)
		  {
			  if(b[i][j]==1){
				  int f=0;
				  for(int k=0;k<m&&f==0;k++){if(a[k][j]==1)f=1;}
				  for(int k=0;k<n&&f==0;k++){if(a[i][k]==1)f=1;}
				  if(f==0)flag=1;
				  }
			 if(flag==1)break;
		  }if(flag==1)break;
	  }
	  if(flag==1)
	  System.out.println("NO");
	  else {System.out.println("YES");
		  for(int i=0;i<m;i++)
			{
				for(int j=0;j<n;j++)
					System.out.print(a[i][j]+" ");
				System.out.println();
			}
		}
	  
	  
}
   
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
}