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
import sys

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

d = defaultdict(int)

ans = 'YES'

if n < k:
    ans = 'NO'
    sys.exit(0)

for c in l:
    d[c] += 1
    if d[c] > k:
        ans = 'NO'
        break

if ans == 'YES':
    print(ans)
    arr = []
    color = 1
    used = {}
    for c in l:
        if color <= k:
            arr.append(color)
            used[c] = used.get(c, [False for _ in range(k + 2)])
            #print('len:%d' % len(used[c]))
            #print('c:'+str(color))
            used[c][color] = True
            color += 1
        else:
            used[c] = used.get(c, [False for _ in range(k + 2)])
            for i in range(1, k + 1):
                #print('i:%d'  % i)
                if used[c][i] == False:
                    arr.append(i)
                    used[c][i] = True
                    break
    print(' '.join(map(str, arr)))
else:
    print(ans)
