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
import os,io
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
s = list(map(int,input().split()))
s1 = {}

for i in range(n):
    if s[i] in s1:
        s1[s[i]].append(i)
    else:
        s1[s[i]] = [i]

s2 = [i for i in s1 if len(s1[i])==1]
ls2 = len(s2)
if ls2%2==1 and ls2==n:
    print("NO")
else:
    ec= 0
    oc = 0
    v = -1
    if ls2%2 == 0:
        print("YES")
        ec = 1
    else:
        f = 0
        for i in s1:
            if len(s1[i])>2:
                v = i
                f = 1
                break
        if f==0:
            print("NO")
        else:
            print("YES")
            oc = 1
    if ec == 1:
        s = ['A']*n
        mid = ls2//2
        for i in s2[mid:]:
            s[s1[i][0]]='B'
        print(''.join(s))
    elif oc == 1:
        s = ['A'] * n
        mid = ls2 // 2
        for i in s2[mid:]:
            s[s1[i][0]] = 'B'
        for i in s1[v][1:]:
            s[i]='B'
        print(''.join(s))