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 _1102B {
    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];
        int[]  freq = new int[5001];
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
             arr[i] = sc.nextInt();
            freq[arr[i]]++;
        }
        int f=0,p=1;
        for (int i = 0; i < 5001; i++) {
            if (freq[i] > k) {
                f = 1;
                break;
            }
            if (freq[i] > 0) {
                for (int j = 0; j < n; j++) {
                    if (arr[j] == i) {
                        if (p > k)
                            p = 1;
                        res[j] = p;
                        p++;
                    }
                }
            }
        }
        if( f==1)
            System.out.println("NO");
        else {
            System.out.println("YES");
            for(int i=0;i<n;i++)
                System.out.print(res[i] + " ");
        }
    }
}