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.*;
import static java.lang.Math.*;

public class SolutionB {
       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][2];
              int max = 0;
              int dict[] = new int[5001];
              for(int i = 0; i < n; i++){
                     a[i][0] = sc.nextInt();
                     a[i][1] = i;
                     dict[a[i][0]]++;
                     max = Math.max(max, dict[a[i][0]]);
              }
              if(max > k){
                     System.out.println("NO");
                     return;
              }
              Arrays.sort(a, new Comparator<int[]>(){
                     public int compare(int i1[], int i2[]){
                            return i1[0] - i2[0];
                     }
              });
              int ans[] = new int[n];
              int c = -1;
              for(int i = 0; i < n; i++){
                     c = ++c % k; 
                     ans[a[i][1]] = c + 1;
              }
              System.out.println("YES");
              for(int curr : ans)
                     System.out.print(curr + " ");
              System.out.println();
	}
}