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 ss
{public static void main(String ar[])
{Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int k=sc.nextInt();
int a[]=new int[n];
int ans[]=new int[n];
HashMap<Integer,Integer>hm=new HashMap<Integer,Integer>();int max=0;
for(int i=0;i<n;i++)
{
    a[i]=sc.nextInt();
    hm.put(a[i],0);
    
}int an=0;
int freq[]=new int[k+1];
for(int i=0;i<n;i++)
{
  int f= hm.get(a[i]);
  if(f==k){an=-1;break;}
  ans[i]=f+1;
  freq[f+1]+=1;
  if(max<ans[i])max=ans[i];
    hm.put(a[i],f+1);
}

if(max<k)
{max++;
   for(int i=0;i<n;i++)
   {  if(max>k)break;
       if(freq[ans[i]]==1)continue;
       else
      {freq[ans[i]]-=1; ans[i]=max++;
   }
    
    
    
}}

if(an==-1||n<k)
System.out.println("NO");
else
{System.out.println("YES");

    for(int j=0;j<n;j++)
    System.out.print(ans[j]+" ");
}
}}