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
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  int n, a[105], b[105]{}, x = 0, y = 0, z = 0;
  char c = 'A', d;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], b[a[i]]++;
  for (int i = 0; i < 101; i++)
    if (b[i] == 1)
      x++;
    else if (b[i] > 2 && !y)
      y = i;
  if (x % 2 == 0 || y) {
    cout << "YES\n";
    for (int i = 0; i < n; i++)
      if (b[a[i]] == 1) {
        cout << c;
        c = char('A' + 'B' - c);
      } else if (a[i] == y && (x & 1)) {
        if (!z) {
          cout << c;
          c = char('A' + 'B' - c);
          d = c, z = 1;
        } else
          cout << d;
      } else
        cout << 'A';
  } else
    cout << "NO\n";
  return 0;
}
