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.BitSet;
import java.util.Scanner;

/**
 * Created with IntelliJ IDEA.
 * User: saveln
 * Date: 11.11.2014
 * Time: 19:02
 */
public class FunctionCounting {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), m = in.nextInt();
        BitSet[] matrix = new BitSet[n], A = new BitSet[n];
        BitSet onesM = new BitSet(m);
        onesM.flip(0, m);
        for (int i = 0; i < n; i++) {
            matrix[i] = new BitSet(m);
            A[i] = new BitSet(m);
            for (int j = 0; j < m; j++) {
                if(in.nextInt() == 0) matrix[i].set(j);
            }
        }

        // zeros
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if(matrix[i].get(j)) {
                    for (int k = 0; k < n; k++) {
                        A[k].set(j);
                    }
                    A[i] = onesM;
                }
            }
        }

        // ones
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if(!matrix[i].get(j)) {
                    boolean can = true;
                    for (int k = 0; k < n; k++) {
                        can &= A[k].get(j);
                    }
                    can &= A[i].equals(onesM);
                    if(can) {
                        System.out.println("NO");
                        return;
                    }
                }
            }
        }

        System.out.println("YES");

        StringBuilder builder = new StringBuilder();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                builder.append(A[i].get(j) ? 0 : 1);
                builder.append(' ');
            }
            builder.append('\n');
        }
        System.out.println(builder.toString());
    }
}