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:
def OutputforEven(timesDic,indexDic,goodKey,numOfsingles):
    list1 = []
    list2 = []
    singles = 0
    for key, times in timesDic.items():
        if times == 1:
            if singles < numOfsingles/2:
                list1.extend(indexDic[key])
                singles += 1
            else:
                list2.extend(indexDic[key])
        else:
            list1.extend(indexDic[key])
    a = 'A'
    b = 'B'
    dict1 = {}
    dict2 = {}
    for elem in list1:
        dict1[elem] = a
    for elem in list2:
        dict2[elem] = b
    dict1.update(dict2)
    list1.extend(list2)
    listall = []
    for index in sorted(list1):
        listall.append(dict1[index])
    print(''.join(listall))
    return

def OutputforOdd(timesDic, indexDic, goodKey,numOfsingles):
    list1 = []
    list2 = []
    singles = 0
    for key, times in timesDic.items():
        if times == 1:
            if singles < numOfsingles//2+1:
                list1.extend(indexDic[key])
                singles += 1
            else:
                list2.extend(indexDic[key])
        elif key == goodKey:
            list1.extend(indexDic[goodKey])
            last = list1.pop()
            list2.append(last)
        else:
            list1.extend(indexDic[key])
    a = 'A'
    b = 'B'
    dict1 = {}
    dict2 = {}
    for elem in list1:
        dict1[elem] = a
    for elem in list2:
        dict2[elem] = b
    dict1.update(dict2)
    list1.extend(list2)
    listall = []
    for index in sorted(list1):
        listall.append(dict1[index])
    print(''.join(listall))
    return 


n = int(input())
arr = input().split()
timesDic = {} 
indexDic = {}
for i in range(len(arr)):
    if arr[i] in timesDic:
        timesDic[arr[i]] += 1
        indexDic[arr[i]].append(i)
    else:
        timesDic[arr[i]] = 1
        indexDic[arr[i]] = [i]

numOfsingles = 0
goodKey = 0.5
for key, times in timesDic.items():
    if times == 1:
        numOfsingles += 1

    if times > 2 and goodKey == 0.5:
        goodKey = key
if numOfsingles & 1:
    if goodKey == 0.5:
        print('NO')
    else:
        print('YES')
        OutputforOdd(timesDic, indexDic, goodKey, numOfsingles)
else:
    print('YES')
    OutputforEven(timesDic,indexDic,goodKey, numOfsingles)