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 subs4 {
    public static void main(String args[]) {
     
     Scanner ip = new Scanner(System.in);
    
     int n = ip.nextInt();
     int k = ip.nextInt();
     
     int[][] arr = new int[n][3];
     
     for(int i = 0; i < n; i++){
        arr[i][0] = ip.nextInt();
        arr[i][1] = i;
     }
    
    Arrays.sort(arr, (a,b) -> a[0] - b[0]);
    
    int col = 1;
    int[] prev = arr[0];
    prev[2] = col;
    int lastCol = col;
    
    for(int i = 1; i < n; i++){
        
        int[] curr = arr[i];
        
        col++;
        if(col > k) col = 1;
        
        if(prev[0] != curr[0]){
            curr[2] = col;
            lastCol = col;
        }
        else{
            if(col == lastCol){
                System.out.println("NO");
                return;
            }
            curr[2] = col;
        }
        prev = curr;   
    }
    
    int[] ans = new int[n];
    
    for(int[] curr : arr)
        ans[curr[1]] = curr[2];
     
    System.out.println("YES");
    
    for(int i = 0; i < n; i++)
        System.out.printf("%d ", ans[i]);
     
    }
}