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>
using namespace std;
int n, a[101], x, y, s[101], f[101];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], s[a[i]]++;
  for (int i = 1; i <= 100; i++)
    if (s[i] == 1)
      x++;
    else if (s[i] > 2)
      y++;
  if (x % 2 == 1 && y == 0)
    cout << "NO";
  else {
    cout << "YES" << endl;
    int j = 1;
    for (int i = 1; i <= x / 2; j++)
      if (s[a[j]] == 1) f[j] = 1, i++;
    if (x % 2 == 1)
      for (int i = 1; i <= n; i++)
        if (s[a[i]] > 2) {
          f[i] = 1;
          break;
        }
    for (int i = 1; i <= n; i++)
      if (f[i] == 1)
        cout << 'A';
      else
        cout << 'B';
  }
  return 0;
}