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;
int const INF = numeric_limits<int>::max();
long long const LLINF = numeric_limits<long long>::max();
double EPS = 1e-9;
long long const N = 105;
long long a[N];
long long ans[N];
map<char, long long> ma;
long long cnt[2];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ma[a[i]]++;
  }
  for (long long i = 0; i < n; i++) {
    if (ma[a[i]] == 1) {
      if (cnt[0] == cnt[1]) {
        ans[i] = 0;
        cnt[0]++;
      } else {
        ans[i] = 1;
        cnt[1]++;
      }
    } else {
      ans[i] = 0;
    }
  }
  bool flag = cnt[0] == cnt[1];
  if (cnt[0] != cnt[1]) {
    for (long long i = 0; i < n; i++) {
      if (ma[a[i]] > 2) {
        ans[i] = 1;
        cnt[1]++;
        flag = true;
        break;
      }
    }
  }
  if (flag) {
    cout << "YES" << '\n';
    for (long long i = 0; i < n; i++) {
      if (ans[i] == 0)
        cout << "A";
      else
        cout << "B";
    }
    cout << '\n';
  } else
    cout << "NO" << '\n';
  return 0;
}