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.*;
public class hero{
 
 public static void main(String[] args) {
  Scanner no=new Scanner(System.in);
  boolean t=true;
  boolean w=false;
  int m=no.nextInt();
  int n=no.nextInt();
  int u=0;
  int arr[][]=new int[m][n];
  int arr1[][]=new int[m][n];
  //int arr2[][]=new int[m][n];
  for(int i=0;i<m;i++){
    for(int j=0;j<n;j++){
      arr1[i][j]=1;
     }
  }
  for(int i=0;i<m;i++){
    for(int j=0;j<n;j++){
      arr[i][j]=no.nextInt();
      if(arr[i][j]==0){
        Arrays.fill(arr1[i],0);
        for(int k=0;k<m;k++){
          arr1[k][j]=0;
        }
      }
      else{
        w=true;
      }
    }
  }
  int o=0;
  outer:for(int i=0;i<m;i++){
    for(int j=0;j<n;j++){
      if(arr[i][j]==1){
        for(int k=0;k<n;k++){
          /*if(arr[i][k]!=1){
            t=false;
            System.out.println("NO");
            break outer;
          }*/
          o=o+arr1[i][k];
        }
        for(int k=0;k<m;k++){
          /*if(arr[k][j]!=1){
            t=false;
            System.out.println("NO");
            break outer;
          }*/
          o=o+arr1[k][j];
        }
        if(o==0){
          t=false;
        }
        o=0;
      }
    }
  }
  //System.out.println(t);
  if(t){
    System.out.println("YES");
    for(int i=0;i<m;i++){
      for(int j=0;j<n;j++){
        System.out.print(arr1[i][j]+" ");
        }
        System.out.println();
      }
  }
  else{
    System.out.println("NO");
  }
}}