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 OrInMatrix {
 public static void main(String[] args){
     Scanner in = new Scanner(System.in);
     int row = in.nextInt();
     int col = in.nextInt();
     int[][] b = new int[row][col];
     int[][] a = new int[row][col];
     for(int i = 0 ; i < row; i++){
         for(int j = 0 ; j < col ; j++){
             a[i][j] = -1;
         }
     }
     for(int i = 0 ; i < row; i++){
         for(int j = 0; j < col; j++){
             b[i][j] = in.nextInt();
             if(b[i][j] == 0){
                 int k = 0;
                 int t = 0;
                 while(k < col || t < row){
                     if(k < col)
                         a[i][k++] = 0;
                     if(t < row)
                         a[t++][j] = 0;
                     
                 }
             }
         }
     }
     for(int i = 0 ; i < row ; i++){
         for(int j = 0 ; j < col ; j++){
                if(b[i][j] == 1){
                    int k = 0;
                    int t = 0;
                    boolean setOne = false;
                    while(k < col || t < row){
                         if(k < col && a[i][k] != 0){
                             a[i][k] = 1;
                             setOne = true;
                         }
                         if(k < col)
                             k++;
                         if(t < row && a[t][j] != 0){
                             a[t++][j] = 1;
                             setOne = true;
                         }
                         if(t < row)
                             t++;
                     }
                    if(!setOne){
                        System.out.println("NO");
                        return;
                    }
                 } 
         }
     }
     System.out.println("YES");
     for(int i = 0 ; i < row ; i++){
         if(i != 0)
         System.out.println();
         for(int j = 0 ; j < col ; j++){
             System.out.print(a[i][j]+" "); 
         }
     }   
 }
}