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.
#!/usr/bin/python3

from collections import namedtuple

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

pair = namedtuple('data', ['elem', 'ind'])
res = [0]*n

arr = [pair(inp[i], i) for i in range(n)]
arr.sort()
flag = True
count = 0
prev = arr[0]
for i in range(k):
    res[arr[i].ind] = i + 1
    if arr[i].elem == prev.elem:
        count += 1
    else:
        count = 1
    prev = arr[i]

for i in range(k, n):
    if arr[i].elem == prev.elem:
        count += 1
        if count > k:
            flag = False
            break
        res[arr[i].ind] = (res[prev.ind] % k) + 1
    else:
        count = 1
        res[arr[i].ind] = 1
    
    prev = arr[i]

if not flag:
    print("NO")
else:
    print("YES")
    print(*res)
    