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())
v = list(map(int, input().split()))
f = [0 for x in range(1, 105)]
for x in v:
    f[x] = f[x] + 1
d = {}
next = 0
for i in range(1, 101):
    if f[i] == 1:
        if next == 0:
            d[i] = 0
        else:
            d[i] = 1
        next = next ^ 1
    elif f[i] == 2:
        d[i] = 1

if next == 1:
    for i in range(1, 101):
        if f[i] >= 3:
            d[i] = 2
            next = next ^ 1
            break
    if next != 0:
        print("NO")
        exit(0)

for i in range(1, 101):
    if i not in d:
        d[i] = 0

print("YES")
sol = []
for x in v:
    if d[x] == 0:
        sol.append('A')
    elif d[x] == 1:
        sol.append('B')
    else:
        sol.append('B')
        d[x] = 0

print(''.join(sol))



