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:
import collections
n=int(input())
a=[int(i) for i in input().split()]
d=collections.Counter(a)
c=0
m=0
for i in a:
    if d[i]==1:
        c+=1
    elif d[i]>=3:
        m+=1
if c%2!=0 and m==0:
    print("NO")
else:
    l=0
    ans=[]
    print("YES")
    for i in range(0,n):
        if d[a[i]]==1 and l%2==0:
            ans.append("A")
            l+=1
        elif d[a[i]]==1 and l%2==1:
            ans.append("B")
            l+=1
        else:
            ans.append("A")
    if c%2==1:
        for i in range(0,n):
            if d[a[i]]>=3:
                ans[i]="B"
                break
    print(*ans,sep="")