You are given an array a consisting of n integer numbers.

You have to color this array in k colors in such a way that: 

  * Each element of the array should be colored in some color; 
  * For each i from 1 to k there should be at least one element colored in the i-th color in the array; 
  * For each i from 1 to k all elements colored in the i-th color should be distinct. 



Obviously, such coloring might be impossible. In this case, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions above. If there are multiple answers, you can print any.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the length of the array a and the number of colors, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5000) — elements of the array a.

Output

If there is no answer, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions described in the problem statement. If there are multiple answers, you can print any.

Examples

Input


4 2
1 2 2 3


Output


YES
1 1 2 2


Input


5 2
3 2 1 2 3


Output


YES
2 1 1 2 1


Input


5 2
2 1 1 2 1


Output


NO

Note

In the first example the answer 2~ 1~ 2~ 1 is also acceptable.

In the second example the answer 1~ 1~ 1~ 2~ 2 is also acceptable.

There exist other acceptable answers for both examples.
import java.util.HashSet;
import java.io.*;
import java.util.*;
public class Final {
		public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int n=sc.nextInt();int k=sc.nextInt();
		TreeSet<Integer> e[]=new TreeSet[5001];
		int a[]=new int[n];int c=1;
		for(int i=0;i<n;i++) {
			a[i]=sc.nextInt();
			if(e[a[i]]==null) {
				e[a[i]]=new TreeSet<Integer>();
				e[a[i]].add(c);
			}else {
				if(e[a[i]].contains(c))
					out:for(int j=1;j<=k;j++) {
						if(!e[a[i]].contains(j)) {
							e[a[i]].add(j);
							break out;
						}	
					}
				else
					e[a[i]].add(c);
			}
			c=(c%k)+1;
		}
		boolean f=true;TreeSet<Integer> tmp;
		for(int i=0;i<n;i++) {
			if(e[a[i]].isEmpty()) {
				f=false;break;
			}else {
				tmp=e[a[i]];
				a[i]=tmp.pollFirst();
			}
		}
		if(f) {
			out.println("YES");
			for(int i=0;i<n;i++)
				out.print(a[i]+" ");
		}else
			out.println("NO");
		
		out.flush();
	}

}