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 sys import stdin
from collections import Counter
from itertools import zip_longest

inp = [stdin.readline().split(' ') for _ in range(2)]
n, k = map(int, inp[0])
a = [int(el) for el in inp[1]]

res = [None] * len(a)

c = Counter(a)
# 1: if one value in a occurs more than k times, its impossible
if any(freq > k for freq in c.values()):
    print("NO")
else:
    print("YES")
    # dict element in a --> last col used
    cols_used = {el: None for el in c.keys()}
    for el in c.keys():
        cols_used[el] = set()
#    last_col_used = 1
#    print(processed
#   XXX wron if len(keys(c)) < k
    for i, el in enumerate(a):
        cand = (i % k) + 1
        #while cand <= processed[el]:
        while cand in cols_used[el]:
            cand += 1
            if cand > k:
                cand = 1
            #print("{}: {} ? ({})".format(el, cand, cols_used))
#            print(cols_used)
#        processed[el] = cand
        cols_used[el].add(cand)
#        if processed[el] > k:
 #           processed[el] = 1
        res[i] = cand
    
    print(" ".join(map(str, res)))
