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 sys
import math
n=int(input())
l=list(map(int,input().split()))
d=dict()
for i in range(n):
    d[l[i]]=0
for i in range(n):
    d[l[i]]+=1
p=0
t=0
for i in range(n):
    if d[l[i]]==1:
        p+=1
    elif d[l[i]]>2:
        t+=1
if p%2!=0 and t==0:
    print("NO")
    sys.exit()
print("YES")
s=""
c=0
t1=0
for i in range(n):
    if d[l[i]]==1 and c>=p//2:
        s=s+"A"
        c+=1
    elif d[l[i]]==1 and c<p//2:
        s=s+"B"
        c+=1
    elif p%2!=0 and d[l[i]]>2 and t1==0:
            s=s+"B"
            t1=1
    else:
        s=s+"A"
print(s)