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(raw_input())
A =map(int, raw_input().split())

dic={}
for x in A:
    if x not in dic:
        dic[x]=1
    else:
        dic[x]+=1

na=0
na2=0
na3=0
for key in dic:
    if dic[key]==1:
        na+=1
    elif dic[key]==2:
        na2+=1
    else:
        na3+=1
if na%2==0:
    print 'YES'
    S=['A' for i in range(n)]
    ct=0
    for i in range(n):
        x=A[i]
        if dic[x]==1 and ct<na/2:
            S[i]='B'
            ct+=1
    print ''.join(S)
    exit()
elif na3>0:
    print 'YES'
    S=['A' for i in range(n)]
    ct=0
    ft=0
    dt=(na-1)/2
    for i in range(n):
        x=A[i]
        if dic[x]==1 and ct<dt:
            S[i]='B'
            ct+=1
        elif dic[x]>2 and ft<1:
            S[i]='B'
            ft+=1
    print ''.join(S)
    exit()
else:
    print 'NO'
    exit()

