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 Main {
 
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        int count=1;
        int m=1;
        int arr[]=new int[n];
        int x[]=new int[5001];
        int y[]=new int[n];
        int EW[][] =new int[k+1][5001];
        for(int i=0;i<n;i++)
            {
               arr[i]=sc.nextInt();
               x[arr[i]]++;
               if(x[arr[i]]>k)
                   count=0;
            }
        if(count==0||k>n)
            System.out.println("NO");
        else
        {
            for(int i=0;i<n;i++)
            {
                if(m<=k)
                {
                    y[i]=m;
                    EW[m][arr[i]]=1;
                    m++;
                }
                else
                {
                    for(int j=1;j<=k;j++)
                    {
                        if(EW[j][arr[i]]==0)
                            {
                               y[i]=j;
                               EW[j][arr[i]]=1;
                               break;
                            }
                    }
                }
            }
            System.out.println("YES");
            for(int i=0;i<n;i++)
                System.out.print(y[i]+" ");
        }
 
    }
 
}
 
	  		 		 	 			 				 		  	 				