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() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> v(n);
  vector<int> freq(101, 0);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    freq[v[i]]++;
  }
  int sum = 0;
  for (int i = 1; i <= 100; ++i)
    if (freq[i] == 1) sum++;
  int must = sum / 2;
  bool found = false;
  if (sum % 2 == 1)
    must++;
  else
    found = true;
  int A_sum = 0;
  vector<int> res(n, 0);
  for (int i = 0; i < n; ++i)
    if (freq[v[i]] == 1) {
      if (A_sum < must)
        A_sum++;
      else
        res[i] = 1;
    }
  if (!found)
    for (int i = 0; i < n; ++i)
      if (freq[v[i]] >= 3) {
        res[i] = 1;
        found = true;
        break;
      }
  if (found) {
    cout << "YES\n";
    for (int i = 0; i < n; ++i)
      if (res[i])
        cout << 'B';
      else
        cout << 'A';
  } else {
    cout << "NO\n";
  }
}
