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 Anudeep Nekkanti
 */
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];
      int [] rows = new int[n];
      int [] cols = new int[m];

      for (int i = 0; i < rows.length; i++) {
        rows[i] = 1;
      }

      for (int i = 0; i < cols.length; i++) {
        cols[i] = 1;
      }

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

      boolean ok = true;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
          if(b[i][j] == 1) {
            boolean found = false;
            for (int k = 0; k < m; k++) {
              if(rows[i] == 1 && cols[k] == 1) found = true;
            }
            for (int k = 0; k < n; k++) {
              if(rows[k] == 1 && cols[j] == 1) found = true;
            }
            if(!found) {
              ok = false;
            }
          }
        }
      }

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

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

        for (int i = 0; i < n; i++) {
          for (int j = 0; j < m; j++) {
            a[i][j] = 1;
            if(rows[i] == 0 || cols[j] == 0) {
              a[i][j] = 0;
            }
            out.printf("%d ", a[i][j]);
          }
          out.println();
        }
      }

    }
}