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=input().split()
n=int(n)
k=int(k)
se=[[] for i in range(k)]
se1=[[] for i in range(k)]
a=list(map(int,input().split()))
#print(a)
#print(se)
flag2=0
if(n<k):
    print('NO')
else:    
    for i in range(n):
        p=i%k
        if a[i] not in se[p]:
            se[p].append(a[i])
            se1[p].append(i)
        else:
            flag=0
            j=0
            while(a[i] in se[j]):
                j+=1
                if(j==k):
                    flag=1
                    break
            if(flag==1):
                print("NO")
                flag2=1
                break
            else:
                se[j].append(a[i])
                se1[j].append(i)
    if(flag2==0):
        print("YES")
        b=[0]*n
        for i in range(k):
            for j in se1[i]:
                b[j]=i+1
        for i in range(n):
            print(b[i],end=" ")