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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class CodeForces486B {

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String s[] = br.readLine().split(" ");

        int m =  Integer.parseInt(s[0]);
        int n =  Integer.parseInt(s[1]);

        int[][] dor = new int[m][n];
        int [][] oor = new int[m][n];
        for(int[] arr:oor) {
            Arrays.fill(arr,-1);
        }
        for(int i=0;i<m;i++) {
            int [] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            for(int j=0;j<n;j++) dor[i][j] = arr[j];
        }

        int flag=0;

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                int k= dor[i][j];
                if(k==0) {
                    for(int l=0;l<n;l++)
                        oor[i][l]=0;
                    for(int p=0;p<m;p++)
                        oor[p][j]=0;
                }
            }
        }

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                if(oor[i][j]==-1) {
                    oor[i][j]=1;
                }
            }
        }

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {

                int or=0;
                for(int l=0;l<n;l++)
                    or=or| oor[i][l];
                for(int l1=0;l1<m;l1++)
                    or=or|oor[l1][j];
                if(or!=dor[i][j]) {
                    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(oor[i][j]+" ");
                }
                System.out.println();
            }
        }
        }

    }