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;
void NO() {
  cout << "NO\n";
  exit(0);
}
int a[100], freq[101];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    freq[a[i]]++;
  }
  int nice = 0, too_much = 0;
  for (int v = 1; v <= 100; v++) {
    if (freq[v] == 1) {
      nice ^= 1;
    } else if (freq[v] >= 3) {
      too_much = v;
    }
  }
  int split = -1;
  if (nice == 1) {
    if (too_much == 0) NO();
    split = too_much;
  }
  cout << "YES\n";
  char other = 'A';
  string ans(n, '*');
  for (int i = 0, id = 0; i < n; i++) {
    if (freq[a[i]] == 1) {
      ans[i] = ('A' + id);
      id ^= 1;
    } else if (a[i] == split) {
      ans[i] = ('A' + id);
      id ^= 1;
      other = 'A' + id;
      split = 0;
    }
  }
  for (char& ch : ans)
    if (ch == '*') ch = other;
  cout << ans << endl;
  return 0;
}