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.*;

public class b {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] arr = new int[n];
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int maxFreq = 0;
        for(int i=0; i<n; i++) {
            arr[i] = sc.nextInt();
            map.put(arr[i], map.containsKey(arr[i]) ? map.get(arr[i])+1 : 1);
            maxFreq = Math.max(maxFreq, map.get(arr[i]));
        }
        if(maxFreq>k) System.out.println("NO");
        else {
            System.out.println("YES");
            int curColor = 1;
            ArrayList<Integer>[] ind = new ArrayList[5050];
            for(int i=0; i<5050; i++) ind[i] = new ArrayList<>();
            for(int i=0; i<n; i++) ind[arr[i]].add(i);
            StringBuilder sb = new StringBuilder();
            int[] out = new int[n];
            for(int i=0; i<5050; i++)
                for(int j=0; j<ind[i].size(); j++){
                    out[ind[i].get(j)] = curColor;
                    curColor++;
                    if(curColor>k) curColor = 1;
                }
            
            for(int i=0; i<n; i++) {
                sb.append(out[i]).append(" ");
            }
            System.out.println(sb.toString().trim());
        }
    }
}
