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.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author thnkndblv
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskB {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();

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

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

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

        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if ( b[i][j] == 1 ) {
                    boolean ok = false;

                    for (int k = 0; k < n; k++) if ( a[k][j] == 1 ) ok = true;
                    for (int k = 0; k < m; k++) if ( a[i][k] == 1 ) ok = true;

                    if (!ok) {
                        out.println("NO");
                        return;
                    }
                }

        out.println("YES");
        for (int [] v : a){
            for (int x : v)
                out.print(x + " ");
            out.println();
        }
    }
}