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;
queue<int> que[107];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    que[t].push(i);
  }
  vector<int> one, more;
  for (int i = 1; i <= 100; ++i) {
    if (que[i].size() == 1) one.push_back(i);
    if (que[i].size() > 2) more.push_back(i);
  }
  int sz = one.size();
  if (sz % 2 == 0) {
    cout << "YES\n";
    string res(n, 'A');
    for (int i = 0; i < sz / 2; ++i) {
      res[que[one[i]].front()] = 'B';
    }
    for (int i = 1; i <= 100; ++i) {
      if (que[i].size() > 1) {
        while (!que[i].empty()) {
          res[que[i].front()] = 'A';
          que[i].pop();
        }
      }
    }
    cout << res << '\n';
  } else {
    if (more.empty()) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      string res(n, 'A');
      for (int i = 0; i < sz / 2 + 1; ++i) {
        res[que[one[i]].front()] = 'A';
      }
      for (int i = sz / 2 + 1; i < sz; ++i) {
        res[que[one[i]].front()] = 'B';
      }
      int last = more.back();
      res[que[last].front()] = 'B';
      que[last].pop();
      while (!que[last].empty()) {
        res[que[last].front()] = 'A';
        que[last].pop();
      }
      for (int i = 1; i <= 100; ++i) {
        if (i != last && que[i].size() != 1) {
          while (!que[i].empty()) {
            res[que[i].front()] = 'A';
            que[i].pop();
          }
        }
      }
      cout << res << '\n';
    }
  }
  return 0;
}
