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.
from collections import defaultdict


if __name__ == "__main__":
    n, k = map(int, input().split())
    nums = list(map(int, input().split()))

    flag = 1
    tag = defaultdict(int)
    ans = [0] * n

    is_exit = False
    for i in range(n):
        tag[nums[i]] += 1
        ans[i] = tag[nums[i]]
        if ans[i] > k:
            is_exit = True
            print('NO')
            break
        elif ans[i] > flag:
            flag = ans[i]
    if not is_exit:
        sign = 1
        while flag != k:
            for i in range(n):
                if ans[i] == sign:
                    break
            for j in range(i+1, n):
                if flag == k:
                    break
                if ans[j] == sign:
                    flag += 1
                    ans[j] = flag
            sign += 1
        print('YES')
        for i in range(n):
            if i < n - 1:
                print(ans[i], end=' ')
            else:
                print(ans[i])
 		  		  		 				 	       	   	