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.HashSet;
import java.util.LinkedList;
import java.util.Scanner;

public class B {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
int o = reader.nextInt();
int n = reader.nextInt();

int[][] matrix = new int[o][n];
int[][] matrix2 = new int[o][n];

for (int i = 0; i < o; i++) {
for (int j = 0; j < n; j++) {
matrix2[i][j] = 1;
}
}

for (int i = 0; i < o; i++) {
for (int j = 0; j < n; j++) {
matrix[i][j] = reader.nextByte();
if (matrix[i][j] == 0) {
for (int k = 0; k < o; k++) {
matrix2[k][j] = 0;
}
for (int k = 0; k < n; k++) {
matrix2[i][k] = 0;
}
}
}
}

boolean flag = false;

HashSet<Integer> list = new HashSet<Integer>();

for (int i = 0; i < o; i++) {
for (int j = 0; j < n; j++) {
if(matrix[i][j] == 1 && flag == false){
for (int k = 0; k < o; k++) {
list.add(matrix2[k][j]);
}
for (int k = 0; k < n; k++) {
list.add(matrix2[i][k]);
}
if(!list.contains(1)){
flag = true;
break;
}else{
list.clear();
}
}
}
if(flag){
break;
}

}

if(flag){
System.out.println("NO");
}else {
System.out.println("YES");
for (int i = 0; i < o; i++) {
for (int j = 0; j < n; j++) {
System.out.print(matrix2[i][j] + " ");
}
System.out.println();
}
}

}
}