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
for _ in range(0,1):
    n=input()
    from collections import  Counter
    lst=map(int,raw_input().split())
    ck=Counter(lst)
    s1=0
    s2=0
    s3=0
    for i in ck:
        if ck[i]==1:
            s1=s1+1
        elif ck[i]==2:
            s2=s2+1
        else:
            s3=s3+1
    if s1==0:
        print "YES"
        ans = ['B'] * (n)
        print "".join(ans)


    elif s1%2==0 or s3>0:
        ans=[]
        print "YES"
        t=[]
        if s1%2==0:
            for i in range(0,n):
                if ck[lst[i]]==1:
                    t.append(i)
                if len(t)>=s1/2:
                    break
        else:
            for i in range(0,n):
                if ck[lst[i]]>=3:
                    t.append(i)
                    break
            if len(t)<(s1+1)/2 :
                for i in range(0,n):
                    if ck[lst[i]]==1:
                        t.append(i)
                    if len(t)>=(s1+1)/2 :
                        break
       # print t


        #print t
        ans=['B']*(n)
        for i in t:
            ans[i]='A'
        print "".join(ans)
    else:
        print "NO"