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 CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  int n, i, p = -1, t;
  int a[101], c[101] = {0};
  char s[101], f = 66;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", a + i);
    c[a[i]]++;
  }
  for (i = 0; i < n; i++) {
    if ((t = c[a[i]]) == 1) {
      if (f == 66)
        f = 65;
      else
        f = 66;
      s[i] = f;
    } else {
      s[i] = 65;
      if (t > 2) p = i;
    }
  }
  if (f == 65) {
    if (p >= 0)
      s[p] = 66;
    else {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  s[n] = 0;
  puts(s);
  return 0;
}