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())
coun={}
a=list(map(int,input().split()))
odin=0
dva=0
tre=0
for ii in range(n):
    i=a[ii]
    if (i not in coun):
        coun[i]=[ii]
        odin+=1
    elif (len(coun[i])==1):
        coun[i].append(ii)
        odin-=1
        dva+=1
    elif (len(coun[i])==2):
        coun[i].append(ii)
        dva-=1
        tre+=1
    else:
        coun[i].append(ii)

first=0
second=0
#print(odin,dva,tre)
if (odin==0):
    print('YES')
    print('A'*n)
elif (odin%2!=0)and(tre==0):
    print('NO')
else:
    print('YES')
    
    if (odin%2==0)and(tre==0):
        mode=1
    elif (odin%2==0)and(tre>0):
        mode=2
    elif (odin%2==1)and(tre>0):
        first_tre=False
        mode=3
    
    for i in range(len(a)):
        if (len(coun[a[i]])==2):
            a[i]='A'
        elif (mode==1):
            #eto vsegda odin
            if (first>second):
                second+=1
                a[i]='B'
            else:
                first+=1
                a[i]='A'
        elif (mode==2):
            if (len(coun[a[i]])>=3):
                a[i]='A'
            else: #eto vsegda odin
                if (first>second):
                    second+=1
                    a[i]='B'
                else:
                    first+=1
                    a[i]='A'
        else: #mode==3
            if (len(coun[a[i]])>=3):
                if (first_tre==False):
                    first_tre=True
                    if (first<second):
                        a[i]='A'
                        last='B'
                        first+=1
                    else:
                        a[i]='B'
                        last='A'
                        second+=1
                else:
                    a[i]=last
            else: #eto vsegda odin
                if (first>second):
                    second+=1
                    a[i]='B'
                else:
                    first+=1
                    a[i]='A'
                    
    print(''.join(a))
if (first!=second):
    int('kill me')
