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 = map(int,input().split())
tee = list(map(int,input().split()))
arr = []
dic, flag = dict(), True
#for i in input.split() :
if (k > n) :
    flag = False
for i in tee :
    temp = int(i)
    if (temp in dic) :
        dic[temp] += 1
    else :
        dic[temp] = 1
    arr.append(temp)
    if (dic[temp] > k) :
        flag = False
newdic, ans = dict(), []
coldic = dict()
temparr = []
if (flag) :
    for j in range(len(arr)) :
        i = arr[j]
        if (i in newdic) :
            newdic[i] += 1
        else :
            newdic[i] = 1
        ans.append(dic[i] - newdic[i] + 1)
        if ((dic[i]-newdic[i]+1) in coldic) :
            coldic[dic[i]-newdic[i]+1] += 1
            temparr.append(j)
        else :
            coldic[dic[i]-newdic[i]+1] = 1
    for iarr in temparr :
        for col in range(1, k+1) :
            if (col not in coldic) :
                ans[iarr] = col
                coldic[col] = 1
                break
    print("YES")
    for j in ans :
        print(j, end=" ")
else :
    print("NO")