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;
const int MX = 105;
int n, a[MX], acu[MX], rep = -1;
vector<int> un;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    acu[a[i]]++;
  }
  for (int i = 0; i < n; i++)
    if (acu[a[i]] == 1)
      un.push_back(i);
    else if (acu[a[i]] > 2)
      rep = i;
  string res(n, 'A');
  if (un.size() % 2 && rep == -1) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  int f = 0;
  for (int x : un) {
    res[x] = 'A' + f;
    f = 1 - f;
  }
  if (f) res[rep] = 'B';
  cout << res << '\n';
  return 0;
}