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
# link: https://codeforces.com/problemset/problem/1051/C

def N():
    print("NO")
def Y():
    print("YES")    

n = int(input())
a = list(map(int, input().split()))
memo = {}
for i in range(n):
    if a[i] not in memo: memo[a[i]] = 1
    else:                memo[a[i]] += 1
if len(memo) == 1:
    for key in memo:
        if memo[key] == 1:
            N()
        elif memo[key] == 2:
            Y()
            print("AB")   
        else:
            Y()
            print("A"*memo[key])
else:
    c1 = 0
    alter = 1 
    found = 1                    
    for key, value in memo.items():
        if value == 1:
            if alter:
                memo[key] = ["A"]
                alter = 0
            else:
                alter = 1
                memo[key] = ["B"]
            c1 += 1                
    if c1 % 2 == 1:
        found = 0
        for key, value in memo.items():
            if type(value) == int and value >= 3:
                found = 1
                memo[key] = ["A"] * value
                memo[key][0] = "B"
                break
    if not found:
        N()
        exit(0)
    Y()            
    for i in range(n):
        if type(memo[a[i]]) == int:
            print("A", end="")
        else:
            print(memo[a[i]][-1], end="")
            memo[a[i]].pop()
