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() {
  int n;
  cin >> n;
  map<int, int> mp;
  std::vector<int> v;
  for (int i = 0; i < n; i++) {
    int val;
    cin >> val;
    v.push_back(val);
    mp[val]++;
  }
  int nice = 0;
  int flag = -1;
  map<int, int>::iterator it = mp.begin();
  for (it; it != mp.end(); it++) {
    if (it->second == 1)
      nice++;
    else if (it->second > 2)
      flag = it->first;
  }
  if (nice % 2 == 0) {
    nice = nice / 2;
    cout << "YES" << endl;
    for (int i = 0; i < v.size(); i++) {
      if (nice && mp[v[i]] == 1) {
        cout << "A";
        nice--;
      } else
        cout << "B";
    }
    cout << endl;
  } else {
    if (flag > -1) {
      nice = nice / 2;
      cout << "YES" << endl;
      for (int i = 0; i < v.size(); i++) {
        if (nice && mp[v[i]] == 1) {
          cout << "A";
          nice--;
        } else if (v[i] == flag) {
          cout << "A";
          flag = -1;
        } else
          cout << "B";
      }
      cout << endl;
    } else {
      cout << "NO\n";
    }
  }
}
