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.Scanner;
public class Solution654 {
    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];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        int[] b = new int[5001];
        int[] c = new int[n];
        for (int i = 0; i < n; i++) {
            b[a[i]]++;
            c[i] = b[a[i]];
        }
        int max = b[0];   //定义次数
        for (int i = 1; i < b.length; i++) {
            if (b[i] > max) {
                max = b[i];
            }
        }
        boolean t = false;
        if (max > k) {
            System.out.print("NO");
        } else if(max==k){
            System.out.println("YES");
            for (int e : c) {
                System.out.print(e + " ");
            }
        } else{
            int cha = k - max;
            boolean flag=false;
            for(int j=1;j<=k;j++){
                flag=false;
                for (int i = 0; i < n; i++) { // 1 2 1 2 1 2 1    1 2 3 4 5 6 7\
                    if (c[i] == j&&flag) {
                        c[i] = k;
                        k--;
                        cha--;
                    }
                    if(c[i] == j)
                        flag=true;
                    if (cha <= 0) {
                        t = true;
                        break;
                    }
                }
            }
            if (t) {
                System.out.println("YES");
                for (int e : c) {
                    System.out.print(e + " ");
                }
            } else {
                System.out.print("NO");
            }
        }

     }
}
	  	  	 	 		   	      	      		