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.
n, k = [int(i) for i in raw_input().split(' ')]
a = [int(i) for i in raw_input().split(' ')]

max1 = 1
dic1 = {}
flag = 0
for i in a:
    if i in dic1:
        if dic1[i] == 0:
            dic1[i] = 1
        if dic1[i] == k:
            flag = 1
            break
        dic1[i] += 1
    else:
        dic1[i] = 0
if flag == 0:
    print 'YES'
    res = ''
    b = [i for i in range(1, k + 1)]
    d = {}
    for i in a:
        if dic1[i] > 0:
            if b:
                tmp = b.pop()
                res += str(tmp) + ' '
                if i in d:
                    d[i].append(tmp)
                else:
                    d[i] = [tmp]
            else:
                j = 1
                if i not in d:
                    d[i] = [j]
                else:
                    while j in d[i]:
                        j += 1
                d[i].append(j)
                res += str(j) + ' '
        else:
            if b:
                res += str(b.pop()) + ' '
            else:
                res += str(k) + ' '

    print res
else:
    print 'NO'
