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())
a=list(map(int,input().split()))
b=[0]*101
for i in a:
    b[i]+=1
k=0
l=0
for i in b:
    if i==1:
        k+=1
    if i>2:
        l+=1
t=False
s=''
if (k%2==0) or (k%2==1 and l>0):
    t=True
if t:
    print('YES')
    if k%2==0:
        k1=0
        for i in a:
            if b[i]==1 and k1<(k//2):
                s+='A'
                k1+=1
            else:
                s+='B'
        print(s)
    else:
        k1=0
        flag=False
        for i in a:
            if b[i]==1 and k1<(k//2):
                s+='A'
                k1+=1
            elif b[i]>2 and flag==False:
                s+='A'
                flag=True
            else:
                s+='B'
        print(s)
else:
    print('NO')
            
