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
def rmap(): return map(int, input().split())
def rlist(): return list(map(int, input().split()))
def rdi(): return int(input())


n = rdi()
line = rlist()
cnt = [0] * 101
once = 0
triple_ind = 0
triple = 0
for elem in line:
    cnt[elem] += 1
    if cnt[elem] == 1:
        once += 1
    elif cnt[elem] == 2:
        once -= 1
    elif cnt[elem] == 3:
        triple = 1
        
if once % 2 == 0:
    print('YES')
    k = 0
    gg = ['A', 'B']
    for elem in line:
        if cnt[elem] == 1:
            print(gg[k], end='')
            k = (k + 1) % 2
        else:
            print('A', end='')
elif triple == 1:
    print('YES')
    k = 0
    gg = ['A', 'B']
    d = 'A'
    for elem in line:
        if cnt[elem] == 1:
            print(gg[k], end='')
            k = (k + 1) % 2
        elif triple == 1 and cnt[elem] >= 3:
            print(gg[k], end='')
            k = (k + 1) % 2
            triple = 0
            d = gg[k]
        elif cnt[elem] >= 3:
            print(d, end='')
        else:
            print('A', end='')
else:  
    print('NO')
