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.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();	
		int k=sc.nextInt();
		int[] a=new int[n];//存编号
		int[] b=new int[n];//存颜色
		int flag=1;
		HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();

		for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextInt();
			if (hm.containsKey(a[i])) {
				hm.put(a[i],hm.get(a[i])+1);
				if (hm.get(a[i])>k) {
					System.out.println("NO");
					return;
					
				}
			}else {
				hm.put(a[i],1);
			}

		}
		
		System.out.println("YES");

		for (Integer integer : hm.keySet()) {
			for (int i = 0; i < a.length; i++) {
				if (a[i]==integer) {
					b[i]=flag;
					flag++;
					if (flag>k) 
						flag=1;
				}
			}
		}

		for (int i = 0; i < b.length; i++) {
			System.out.print(b[i]+" ");
		}
	}

}
   					 	   		   		 					 				