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, b = 0;
  cin >> n;
  vector<int> s;
  string ch;
  vector<int> nice;
  vector<int>::iterator indice;
  int tab[101];
  int e;
  for (int i = 0; i < n; i++) {
    cin >> e;
    s.push_back(e);
    ch.push_back('A');
  }
  for (int i = 0; i <= 100; i++) tab[i] = 0;
  for (int i = 0; i < n; i++) {
    tab[s[i]]++;
  }
  for (int j = 0; j <= 100; j++) {
    if (tab[j] == 1) nice.push_back(j);
  }
  if (nice.size() % 2 == 0)
    cout << "YES" << endl;
  else {
    for (int k = 0; k <= 100; k++) {
      if (tab[k] > 2) {
        b = 1;
        nice.push_back(k);
        break;
      }
    }
    if (b == 0) {
      cout << "NO" << endl;
      return 0;
    } else
      cout << "YES" << endl;
  }
  int k = 0;
  for (int i = 0; i < nice.size(); i++) {
    indice = find(s.begin(), s.end(), nice[i]);
    if (k == 0) {
      ch[indice - s.begin()] = 'A';
      k++;
    } else {
      k--;
      ch[indice - s.begin()] = 'B';
    }
  }
  cout << ch;
  return 0;
}
