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.*;
import java.util.*;

public class B {
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		int[][] arr = new int[n][m];
		
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < m; j++) {
				arr[i][j] = sc.nextInt();
			}
		}
		
		int[][] ans = new int[n][m];
		for(int e[] : ans) {
			Arrays.fill(e, 1);
		}
		
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < m; j++) {
				if(arr[i][j] == 0) {
					makeZero(ans,i,j);
				}
			}
		}
		boolean ok = false;
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < m; j++) {
				if(arr[i][j] == 1) {
					if(!valid(ans,i,j)) {
						System.out.println("NO");
						ok = true;
						break;
					}
				}
			}
			if(ok) {
				break;
			}
			
		}
		if(!ok) {
			System.out.println("YES");
			for(int i = 0; i < n; i++) {
				for(int j = 0; j < m; j++) {
					System.out.print(ans[i][j] + " ");
				}
				System.out.println();
			}
		}
		
		
	} 
	
	private static boolean valid(int[][] ans, int i, int j) {
		if(ans[i][j] == 1) {
			return true;
		}
	
		for(int x = 0; x < ans[0].length; x++) {
			if(ans[i][x] == 1) {
				return true;
			}
		}
		
		for(int x = 0; x < ans.length; x++) {
			if(ans[x][j] == 1) {
				return true;
			}
		}
		return false;
	}

	private static void makeZero(int[][] ans, int i, int j) {
		ans[i][j] = 0;
		
		for(int x = 0; x < ans[0].length; x++) {
			ans[i][x] = 0;
		}
		
		for(int x = 0; x < ans.length; x++) {
			ans[x][j] = 0;
		}
	}









	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	
	
	
	static int gcd(int a, int b) {
		return a%b == 0 ? b : gcd(b,a%b);
	}
	
	static boolean[] sieve(int n) {
		
		boolean isPrime[] = new boolean[n+1];
		for(int i = 2; i <= n; i++) {
			if(isPrime[i]) continue;
			for(int j = 2*i; j <= n; j+=i) {
				isPrime[j] = true;
			}
		
		}
		return isPrime;
		
	}
}