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.*;
import java.math.*;

public class ORinMatrix {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        final int m = sc.nextByte();
        final int n = sc.nextByte();

        byte[][] b = new byte[m][n];

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

        byte[][] a = new byte[m][n];
        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 row = 0; row < m; row++) {
                        a[row][j] = 0;
                    }

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

        boolean valid = true;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (b[i][j] == 1) {
                    byte check = 0;

                    for (int row = 0; row < m; row++) {
                        check |= a[row][j];
                    }

                    for (int col = 0; col < n; col++) {
                        check |= a[i][col];
                    }

                    if (valid && check == 0)
                        valid = false;
                }
            }
        }
        
        if (valid) {
            System.out.println("YES");
            for (int i = 0; i < m; i++) {
                String row = "";
                for (int j = 0; j < n-1; j++) {
                    row += Byte.toString(a[i][j]) + " ";
                }
                row += Byte.toString(a[i][n-1]);
                System.out.println(row);
            }
        } else {
            System.out.println("NO");
        }
    }
}