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;
vector<pair<int, int> > v, v2;
int main() {
  std::ios::sync_with_stdio(false);
  int T;
  T = 1;
  while (T--) {
    int n, i, j;
    cin >> n;
    int ar[n + 1];
    char ans[n + 1];
    for (i = 1; i <= n; ++i) {
      ans[i] = 'A';
      cin >> ar[i];
      v.push_back({ar[i], i});
    }
    sort(v.begin(), v.end());
    int ct = 0, ct2 = 0;
    int st = 0;
    for (i = 1; i < v.size(); ++i) {
      if (v[i].first != v[i - 1].first) {
        if (st != i - 1) {
          if (st != i - 2) v2.push_back(v[i - 1]);
        } else {
          if (ct > ct2) {
            ans[v[i - 1].second] = 'B';
            ++ct2;
          } else {
            ans[v[i - 1].second] = 'A';
            ++ct;
          }
        }
        st = i;
      }
    }
    if (st != i - 1) {
      if (st != i - 2) v2.push_back(v[i - 1]);
    } else {
      if (ct > ct2) {
        ans[v[i - 1].second] = 'B';
        ++ct2;
      } else {
        ans[v[i - 1].second] = 'A';
        ++ct;
      }
    }
    if (ct == ct2) {
      cout << "YES" << endl;
      for (i = 1; i <= n; ++i) cout << ans[i];
      cout << endl;
    } else if (v2.size() != 0) {
      ans[v2[0].second] = 'B';
      cout << "YES" << endl;
      for (i = 1; i <= n; ++i) cout << ans[i];
      cout << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}