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 collections import Counter


n = int(input())
s = list(map(int, input().split()))

count = Counter(s)
count_1 = 0
count_2 = 0
for i in range(n):
    if count[s[i]] == 1:
        count_1 += 1
    if count[s[i]] == 2:
        count_2 += 1

if count_1 % 2 == 1 and count_1 + count_2 == n:
    print("NO")
else:
    print("YES")
    res = []
    cls = ['A', 'B']
    cur = 0
    if count_1 % 2 == 0:
        for i in range(n):
            if count[s[i]] == 1:
                res.append(cls[cur % 2])
                cur += 1
            else:
                res.append('B')
    else:
        need = True
        for i in range(n):
            if count[s[i]] == 1:
                res.append(cls[cur % 2])
                cur += 1
            elif need and count[s[i]] > 2:
                res.append('B')
                need = False
            else:
                res.append('A')
    print(''.join(res))
		 		  	  		 	 		  	   		 	 	 	