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 main() {
  long long n;
  cin >> n;
  vector<long long> num(n);
  unordered_map<long long, long long> mp;
  long long c = 0;
  long long idx = -1;
  for (int i = 0; i < n; i++) {
    cin >> num[i];
    mp[num[i]]++;
    if (mp[num[i]] == 1) c++;
    if (mp[num[i]] == 2) c--;
    if (mp[num[i]] >= 3) idx = i;
  }
  if (c % 2 && idx == -1)
    cout << "NO" << endl;
  else {
    if (c % 2 == 0) {
      cout << "YES" << endl;
      char v = 'A';
      for (int i = 0; i < n; i++) {
        if (mp[num[i]] == 1) {
          cout << v;
          v = (v == 'A') ? 'B' : 'A';
        } else
          cout << 'B';
      }
      cout << endl;
    } else {
      cout << "YES" << endl;
      char v = 'A';
      for (int i = 0; i < n; i++) {
        if (mp[num[i]] == 1) {
          cout << v;
          v = (v == 'A') ? 'B' : 'A';
        } else if (i == idx)
          cout << 'B';
        else
          cout << 'A';
      }
      cout << endl;
    }
  }
}