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;
const long long INF = 2e9 + 5, N = 2e5 + 5, mod = 1e9 + 7, M = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int a[n + 1];
  map<int, int> cnt;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int must = 0;
  for (int i = 1; i <= 100; i++) {
    if (cnt[i] == 1) {
      must++;
    }
  }
  if (must % 2 == 0) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) {
      if (cnt[a[i]] == 1) {
        if (must % 2)
          cout << 'A';
        else
          cout << 'B';
        must--;
      } else {
        cout << 'A';
      }
    }
    cout << endl;
    return 0;
  }
  bool can = false;
  int num, nam;
  for (int i = 1; i <= 100; i++) {
    if (cnt[i] >= 3) {
      can = true;
      num = i;
      nam = cnt[i];
    }
  }
  if (!can) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]] == 1) {
      if (must % 2)
        cout << 'A';
      else
        cout << 'B';
      must--;
    } else if (a[i] == num) {
      if (nam == 1)
        cout << 'B';
      else
        cout << 'A';
      nam--;
    } else {
      cout << 'A';
    }
  }
}
