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
from sys import stdin, stdout

n = int(stdin.readline())
vl = list(map(int, stdin.readline().split()))
sze = 110
a, b = [], []
used = [[] for i in range(sze)]
cnta, cntb = 0, 0

for i in range(n):
    used[vl[i]].append(i)
    
used.sort(key = lambda x: len(x))

for i in range(sze):
    if len(used[i]) == 1:
        if cnta < cntb:
            a.append(used[i][0])
            cnta += 1
        else:
            b.append(used[i][0])
            cntb += 1
            
    elif len(used[i]) == 2:
        a.append(used[i][0])
        b.append(used[i][1])
        cnta += 1
        cntb += 1
    elif len(used[i]) > 2:
        if cnta == cntb:
            for j in range(len(used[i])):
                a.append(used[i][j])
        elif cnta < cntb:
            cnta += 1
            a.append(used[i][0])
            
            for j in range(1, len(used[i])):
                b.append(used[i][j])   
        else:
            cntb += 1
            b.append(used[i][0])
            
            for j in range(1, len(used[i])):
                a.append(used[i][j]) 
                
                
if cnta == cntb:
    stdout.write('YES\n')
    ans = ['' for i in range(n)]
    
    for v in a:
        ans[v] = 'A'
    
    for v in b:
        ans[v] = 'B'
        
    stdout.write(''.join(ans))
else:
    stdout.write('NO\n')