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 in = new Scanner(System.in);
        int n = in.nextInt(), k = in.nextInt(), max = 0, sum = 0;
        int[] a = new int[n], f = new int[5001];
        for (int i = 0; i < n; i++)
            max = Math.max(max, ++f[a[i] = in.nextInt()]);
        for (int i : a)
            sum += f[i] > 0 ? 1 : 0;
        
        boolean[] v = new boolean[5001];
        if (sum >= k && max <= k)
        {
            System.out.println("YES");
            int at = k;
            for (int i : a)
            {
                if (!(at <= 0 || v[i]))
                {
                    int tmp = f[i];
                    if (tmp >= at);
                    else 
                    { 
                        f[i] = at; 
                        v[i] = true;
                    }
                    at -= tmp;
                }
                System.out.print((f[i]--)+" ");
            }
        }
        
        else
            System.out.println("NO");
        in.close();
    }
}