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() {
  int n;
  cin >> n;
  vector<int> v;
  map<int, int> m;
  int a;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    v.push_back(a);
    if (((m).find(a) != (m).end())) {
      m[a]++;
    } else {
      m[a] = 1;
    }
  }
  int c = 0;
  int f = -1;
  for (auto &el : m) {
    if (el.second == 1) {
      ++c;
    }
    if (el.second > 2) {
      f = el.first;
    }
  }
  if (c % 2 && f == -1) {
    cout << "NO";
    { return 0; };
  }
  cout << "YES" << endl;
  bool f1 = false, f2 = false;
  for (int i = 0; i < n; ++i) {
    int u = m[v[i]];
    if (u == 1) {
      cout << (f1 ? 'A' : 'B');
      f1 = !f1;
    } else if (u == 2) {
      cout << 'A';
    } else if (v[i] == f && c % 2) {
      cout << (f2 ? 'B' : 'A');
      f2 = true;
    } else {
      cout << 'B';
    }
  }
  { return 0; };
}