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())
s = list(map(int,input().split()))

def getstr(a):
    ans = ""
    for i in a:
        ans += i
    return ans

count = [0]*101
ans = []

for i in range(n):
	count[s[i]] += 1

a = 0
b = 0

for i in range(101):
	if (count[i] > 0):
		if (count[i] == 1):
			a += 1
		elif (count[i] > 2):
			b += 1

if (a%2 == 0):
	idx = 0
	for i in range(n):
		if (count[s[i]] == 1):
			if (idx == 1):
				ans.append("A")
			else:
				ans.append("B")
			idx = 1 - idx
		else:
			ans.append("A")
	print("YES")
	print(getstr(ans))
elif (a%2 == 1 and b>0):
	ele = -1
	for i in range(n):
	    if (count[s[i]] > 2):
	        ele = s[i]
	        break
	done = False
	idx = 0
	for i in range(n):
		if (count[s[i]] == 1):
			if (idx == 1):
				ans.append("A")
			else:
				ans.append("B")
			idx = 1 - idx
		else:
		    if (s[i] == ele):
		        if (not done):
		            ans.append("A")
		            done = True
		        else:
		            ans.append("B")
		    else:
			    ans.append("A")
	print("YES")
	print(getstr(ans))
else:
	print("NO")