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 maxn = (int)(5e5 + 123);
int n;
int a[1001];
int used[1001];
char ch[1001];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    used[a[i]]++;
  }
  int ans = 0, ans2 = 0;
  for (int i = 1; i <= 100; ++i) {
    if (used[i] == 1) ans++;
    if (used[i] > 2) ans2++;
  }
  if (ans % 2 == 0) {
    cout << "YES\n";
    int cnt = ans;
    for (int i = 1; i <= n; ++i) {
      if (used[a[i]] == 1) {
        if (cnt > ans / 2) {
          cout << 'A';
        } else {
          cout << 'B';
        }
        --cnt;
      } else {
        cout << 'A';
      }
    }
  } else if (ans2) {
    cout << "YES\n";
    int cnt = ans;
    bool was = 0;
    for (int i = 1; i <= n; ++i) {
      if (used[a[i]] == 1) {
        if (cnt > ans / 2) {
          cout << 'A';
        } else {
          cout << 'B';
        }
        --cnt;
      } else {
        if (used[a[i]] > 2 && !was) {
          cout << 'B';
          was = 1;
        } else {
          cout << 'A';
        }
      }
    }
  } else {
    cout << "NO";
  }
  return 0;
}