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 B {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        int m = input.nextInt();
        int n = input.nextInt();

        int[][] b = new int[m][n];
        int[][] a = new int[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                b[i][j] = input.nextInt();
            }
        }

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

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (b[i][j] == 0) {
                    for (int l = 0; l < n; l++) {
                        a[i][l] = 0;
                    }
                    for (int l = 0; l < m; l++) {
                        a[l][j] = 0;
                    }
                }
            }
        }

        boolean check = true;
        int[][] c = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int orVal = 0;
                for (int l = 0; l < n; l++) {
                    if (a[i][l] == 1) {
                        orVal = 1;
                    }
                }

                for (int l = 0; l < m; l++) {
                    if (a[l][j] == 1) {
                        orVal = 1;
                    }
                }

                c[i][j] = orVal;
            }
        }
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (b[i][j] != c[i][j]) {
                    
                    check = false;
                }
            }
        }

        if (check == true) {
            System.out.println("YES");
            printArray(a);
        } else {
            System.out.println("NO");
        }

    }

    static void printArray(int a[][]) {
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < a[i].length; j++) {
                System.out.print(a[i][j] + " ");
            }
            System.out.println();
        }
    }

}