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())
a=[int(i) for i in input().split()]
da={}
o=list('Z'*n)
for i in a:
    if i in da:
        da[i]+=1
    else:
        da[i]=1
sc=0
dc=0
mc=0
scl=[]
dcl=[]
mcl=[]
for i in da:
    if da[i]==1:
        sc+=1
        scl.append(i)
    elif da[i]==2:
        dc+=1
        dcl.append(i)
    else:
        mc+=1
        mcl.append(i)
#print(sc,dc,mc,scl,dcl,mcl)
for i in dcl:
    x=2
    for j in range(n):
        if a[j]==i:
            o[j]='A'
            x-=1
        if x==0:
            break
pc=0
for i in scl:
        j=a.index(i)
        if pc%2==0:
            o[j]='A'
        else:
            o[j]='B'
        pc+=1
f=1
if sc==0:
    for i in range(n):
        o[i]='A'
elif sc%2==0:
    for i in mcl:
        x=da[i]
        for j in range(n):
            if a[j]==i:
                o[j]='A'
                x-=1
            if x==0:
                break
elif sc%2==1:
    if len(mcl)==0:
        f=0
    else:
        pc=0
        for i in mcl:
            x=da[i]
            for j in range(n):
                if a[j]==i:
                    if pc==0:
                        o[j]='B'
                        pc=1
                    else:
                        o[j]='A'
                    x-=1
                if x==0:
                    break
if f==0:
    print('NO')
else:
    print('YES')
    print(''.join(o))
    