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.
def main():
  from collections import deque
  n, k = map(int, input().split(" "))
  a = [int(x) for x in input().strip().split(" ")]
  counts = {}
  for i, elem in enumerate(a):
    if elem not in counts:
      counts[elem] = deque()
    counts[elem].append(i)
    if len(counts[elem]) > k:
      print("NO")
      return

  color_start = 0
  for elem in counts:
    indices = counts[elem]
    color_offset = 0
    while len(indices) > 0:
      color = ((color_start + color_offset) % k) + 1
      idx = indices.popleft()
      a[idx] = color
      color_offset += 1

    color_start = ((color_start + color_offset) % k)
  print("YES")
  print(" ".join([str(x) for x in a]))
main()
