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
#codeforces_1015C
#please, somebody kill me
gi = lambda : list(map(int,input().split()))
n, = gi()
l = gi()
occ = [0]*101
for e in l:
	occ[e] += 1
niceable,niceable2 = 0,0
for k in range(1,101):
	if occ[k] in (1,2):
		niceable += occ[k]
	elif occ[k] > 0:
		niceable2 += 1
if niceable2 == 0 and niceable%2 :
	print("NO")
	exit()
print("YES")
cur1, cur2, index = 0,0,0
if (niceable2+niceable)%2:
	for e in l:
		if occ[e] > 2:
			occ[e] = -2
			index = e
			break;
A = []
B = []
for k in range(1,101):
	if occ[k] == 1:
		if cur1%2:
			A.append(k)
		else:
			B.append(k)
		cur1 += 1
	elif occ[k] == 2:
		A.append(k)
		B.append(k)
if len(A) > len(B):
	cur = 1
else:
	cur = 0
for k in range(1,101):
	if occ[k] > 2:
		if cur%2 == 0:
			A.append(k)
			B += [k] * (occ[k] - 1)
		else:
			B.append(k)
			A += [k] * (occ[k] - 1)
		cur += 1
A += [index] * l.count(index)
for k in range(n):
	if l[k] in A:
		print("A",end="")
		A.remove(l[k])
	else:
		print("B",end="")
		B.remove(l[k])