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;

/**
 * Created by constanzafierro on 17-06-16.
 */
public class Problem486B {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] line = br.readLine().split(" ");
        int m = Integer.parseInt(line[0]);
        int n = Integer.parseInt(line[1]);
        int[][] b = new int[m][n];
        for(int i = 0; i < m; i++){
            line = br.readLine().split(" ");
            for(int j = 0; j < n; j++)
                b[i][j] = Integer.parseInt(line[j]);
        }
        // making A
        int[][] a = new int[m+1][n+1];
        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 u = 0; u <= n; u++)
                        a[i+1][u] = 0;
                    for(int u = 0; u <= m; u++)
                        a[u][j+1] = 0;
                }
            }
        }

        // save OR of rows
        for(int i = 1; i <= m; i++) {
            int j = 1;
            while(j <= n && a[i][j] == 0){
                j++;
            }
            if(j > n){
                a[i][0] = 0;
            }
        }
        // save OR of columns
        for(int j = 1; j <= n; j++) {
            int i = 1;
            while(i <= m && a[i][j] == 0){
                i++;
            }
            if(i > m){
                a[0][j] = 0;
            }
        }

        for(int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (b[i][j] == 1){
                    if(a[i+1][0] == 0 && a[0][j+1] == 0){
                        System.out.println("NO");
                        return;
                    }
                }
            }
        }

        System.out.println("YES");
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++)
                sb.append(a[i][j]).append(" ");
            sb.append('\n');
        }
        System.out.println(sb.toString());
    }
}