Problem: 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

Solution (in PYTHON3):

Solution:
n=int(input())
l=list(map(int,input().split()))
d={}
for i in l:
    try:
        d[i]+=1
    except:
        d[i]=1
x=list(d.values())
if x.count(1)%2==0:
    print('YES')
    c=0
    ans=['']*n
    for i in range(n):
        if d[l[i]]==1:
            if c:
                ans[i]='B'
            else:
                ans[i]='A'
            c=c^1
        else:
            ans[i]='A'
    print("".join(ans))
    
else:
    if x.count(1)+x.count(2)<len(x):
        print('YES')
        flag=1
        c=0
        ans=['']*n
        for i in range(n):
            if d[l[i]]==1:
                if c:
                    ans[i]='B'
                else:
                    ans[i]='A'
                c=c^1
            elif d[l[i]]==2:
                ans[i]='A'
            else:
                if flag:
                    ans[i]='B'
                    flag=0
                else:
                    ans[i]='A'
                
        print("".join(ans))
    else:
        print('NO')