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.
'''
Array K-Colouring. (1400 Rating).

In these YES OR NO type problems, it is often easier to solve one than the other.
Here, when do I know that no colouring is possible?
Well, since I cannot use the same colour for elements with the same value, any element that appears more than k times automatically ruins the colouring. 
Therefore, this should be a good first check.
Count the # of occurrences of each element. If any element occurs more than k times, then a colouring is impossible because we will violate rule 3.

Now assume that this check succeeds. Are we guaranteed a Yes answer? The answer is yes we are.
Since k<=n, we can guarantee that we will use each colour.
Just go through each element in the Counter. For elements with the same value, keep assigning them a new value. 
Once we hit the top value of k, we just reset it back to 1 and continue colouring.

Runtime is O(n).

'''
import collections
def solve(arr, k):
    counts = collections.Counter(arr)
    indices_of_elem = collections.defaultdict(list)
    if counts.most_common(1)[0][1] > k:
        return ("NO", [])
    for i,elem in enumerate(arr):
        indices_of_elem[elem].append(i)
    result = [0] * len(arr)
    curK = 1
    for key in indices_of_elem.keys():
        for index in indices_of_elem[key]:
            result[index] = curK
            curK = (curK  % k) + 1
    
    return ('YES', result)

n, k = map(int, input().split(' '))
arr = list(map(int, input().split(' ')))

possible, result = solve(arr, k)
print(possible)
if possible == 'YES': 
    print(*result)