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;
set<int> st[2];
map<int, int> mp;
int ar[110];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    mp[ar[i]]++;
  }
  int idx = 0;
  map<int, int>::iterator it1;
  for (it1 = mp.begin(); it1 != mp.end(); it1++) {
    if (it1->second == 1) {
      st[idx].insert(it1->first);
      idx = !idx;
    }
  }
  if (st[0].size() == st[1].size()) {
    string ans(n, 'A');
    for (int i = 0; i < n; i++) {
      if (st[1].count(ar[i])) ans[i] = 'B';
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << ans[i];
    return 0;
  } else {
    string res(n, 'A');
    for (int i = 0; i < n; i++) {
      if (st[1].count(ar[i])) {
        res[i] = 'B';
      }
    }
    int id = -1;
    map<int, int>::iterator it;
    for (it = mp.begin(); it != mp.end(); it++) {
      if (it->second >= 3) {
        id = it->first;
        break;
      }
    }
    if (id == -1) {
      cout << "NO";
      return 0;
    }
    bool flag = false;
    for (int i = 0; i < n; i++) {
      if (ar[i] == id) {
        if (!flag) {
          flag = true;
          res[i] = 'B';
        }
      }
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << res[i];
    }
    return 0;
  }
  return 0;
}
