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 java.io.*;
import java.lang.*;
public final class queue{
 	public static int[] fre=new int[5000];
	public static void main(String args[])
	{
	Scanner sc=new Scanner(System.in);
	int n,c,i,l,arr[],max=0,k;
	n=sc.nextInt();
	c=sc.nextInt();
	int[][] pp=new int[5000][c];
	//System.out.print(pp[8][9]);
	arr=new int[n];
	for(i=0;i<5000;i++)
		fre[i]=0;
		for(i=0;i<n;i++)
		{
			arr[i]=sc.nextInt();
			fre[arr[i]-1]++;
			if(fre[arr[i]-1]>max)
				max=fre[arr[i]-1];
		}	
		if(max>c)
		System.out.println("NO");
		else
		{
			System.out.println("YES");
			k=l=0;
				for(i=0;i<n;i++)
				{
					while(true)
						{
							if(pp[arr[i]-1][l]!=0)
							l=(l+1)%c;
							else
								{pp[arr[i]-1][l]=1;
								max=l+1;
								System.out.print(max+" ");
								break;}
						}
						l=(k+1)%c;
						k=(k+1)%c;
				}
		}
	}
}