Vasya has a multiset s consisting of n integer numbers. Vasya calls some number x nice if it appears in the multiset exactly once. For example, multiset \{1, 1, 2, 3, 3, 3, 4\} contains nice numbers 2 and 4.

Vasya wants to split multiset s into two multisets a and b (one of which may be empty) in such a way that the quantity of nice numbers in multiset a would be the same as the quantity of nice numbers in multiset b (the quantity of numbers to appear exactly once in multiset a and the quantity of numbers to appear exactly once in multiset b).

Input

The first line contains a single integer n~(2 ≤ n ≤ 100).

The second line contains n integers s_1, s_2, ... s_n~(1 ≤ s_i ≤ 100) — the multiset s.

Output

If there exists no split of s to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of n characters. i-th character should be equal to 'A' if the i-th element of multiset s goes to multiset a and 'B' if if the i-th element of multiset s goes to multiset b. Elements are numbered from 1 to n in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples

Input

4
3 5 7 1


Output

YES
BABA


Input

3
3 5 1


Output

NO
n = int(input())
from collections import Counter

m=list(map(int,input().split()))
d=Counter(m)
k=0
r=0
for c in d:
    if d[c]==1:
        k+=1
    if d[c]>2:
        r+=1
if r==0 and k%2:
    print('NO')
else:
    s=''
    if k%2:
        l=(k-1)//2 
        j=1
    else:
        l=k//2
        j=0
    i=0
    h=set()
    for i in range(n):
        if d[m[i]]==1:
            if l>0:
                l-=1
                s+='A'
            else:
                s+='B'
        elif d[m[i]]>2 and j>0:
            if m[i] not in h:
                s+='A'
                h.add(m[i])
                j-=1
            else:
                s+='B'
        else:
            s+='B'
    print('YES')
    print(s)

        
            
            
        
    
    