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
n=int(input())
l=list(map(int,input().split()))
ans=[0]*n
check=[0]*(101) 
for i in range(n):
    check[l[i]]+=1 
once=0 
for i in range(101):
    if check[i]==1:
        once+=1 
keep=1   
if once%2==0:
    print("YES")
    for i in range(n):
        if check[l[i]]==1 and keep<=once//2:
            ans[i]='A'
            keep+=1
        else:
            ans[i]='B'
    print(''.join(ans))    
else:
    if max(check)<=2:
        print("NO")
    else:
        keep=1
        print("YES")
        convert=0
        for i in range(1,101):
            if check[i]>2:
                convert=i 
                break
        flag=0   
        for i in range(n):
            if check[l[i]]==1 and keep<=once//2:
                ans[i]='A'
                keep+=1
            elif l[i]==convert and flag==0:
                ans[i]='A'
                flag=1
            else:
                ans[i]='B'
        print(''.join(ans))            
                
                