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 MAX = 1e5;
set<long long> f, s;
vector<long long> vec, present[MAX];
long long N, lol, arr[MAX], cnt[MAX];
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (long long A = 1; A <= N; A++) {
    cin >> arr[A];
    cnt[arr[A]]++;
    if (cnt[arr[A]] > 1) lol = 1;
  }
  for (long long A = 1; A <= N; A++)
    if (cnt[arr[A]] == 1)
      vec.push_back(A);
    else
      present[arr[A]].push_back(A);
  if (!vec.empty()) {
    f.insert(vec[0]);
    for (long long A = 1; A < vec.size(); A++) {
      if (A % 2)
        s.insert(vec[A]);
      else
        f.insert(vec[A]);
    }
  }
  if (!(vec.size() % 2)) {
    for (long long A = 1; A <= 100; A++) {
      if (present[A].empty()) continue;
      for (long long B = 0; B < present[A].size(); B++) s.insert(present[A][B]);
    }
  } else {
    if (!lol) return cout << "NO", 0;
    long long ok = 1;
    for (long long A = 1; A <= 100; A++) {
      if (present[A].empty())
        continue;
      else if (ok and present[A].size() > 2) {
        ok = 0;
        s.insert(present[A][0]);
        for (long long B = 1; B < present[A].size(); B++)
          f.insert(present[A][B]);
      } else {
        for (long long B = 0; B < present[A].size(); B++)
          s.insert(present[A][B]);
      }
    }
    if (ok) return cout << "NO", 0;
  }
  cout << "YES\n";
  for (long long A = 1; A <= N; A++) {
    if (f.count(A))
      cout << "A";
    else if (s.count(A))
      cout << "B";
    else
      assert(false);
  }
  return 0;
}
