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.ArrayList;
import java.util.*;
public class penshe {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int result[];
		result = new int[n];
		int q[];
		q = new int[n];
		for(int i=0;i<n;i++) {
			q[i] = in.nextInt();
		}
		in.close();
		Map<Integer, List<Integer>> util = new HashMap<>();
		int c=1;
		for(int i=0;i<n;i++) {
			if(util.containsKey(q[i])){
				if(c>k) {c=1;}
				int count=0;
				while(util.get(q[i]).contains(c)) {
					count++;
					if(count>k) {System.out.println("NO");return;}
					c++;
					if(c>k) {c=1;}
				}
				util.get(q[i]).add(c);
				result[i]=c;
				c++;
			}
			else {
				if(c>k) {c = 1;}
				List<Integer> as = new ArrayList<>();
				as.add(c);
				util.put(q[i], as);
				result[i] = c;
				c++;
			}
		}
		System.out.println("YES");
		for(int i=0;i<n;i++) {
			System.out.print(result[i]+" ");
		}
	}
}
			     		 		 	  		   	 				 		