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(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, a[101] = {0};
    cin >> n;
    long long x;
    long long v[n];
    vector<long long int> o, one;
    for (long long int i = 0; i < n; i++) {
      cin >> v[i];
      a[v[i]]++;
    }
    for (long long int i = 1; i < 101; i++) {
      if (a[i] == 1) {
        one.push_back(i);
      } else if (a[i] > 2)
        o.push_back(i);
    }
    if (((long long int)one.size() % 2) && (o.empty())) {
      cout << "NO";
    } else {
      vector<char> s(n, 'A');
      for (long long int i = 0; i < (long long int)one.size() / 2; i++) {
        for (long long int j = 0; j < n; j++) {
          if (v[j] == one[i]) s[j] = 'B';
        }
      }
      if ((long long int)one.size() % 2) {
        for (long long int j = 0; j < n; j++) {
          if (v[j] == o[0]) {
            s[j] = 'B';
            break;
          }
        }
      }
      cout << "YES\n";
      for (long long int i = 0; i < n; i++) cout << s[i];
    }
  }
  return 0;
}
