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;
long long const mod = 2;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modInv(long long x) { return power(x, mod - 2, mod) % mod; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long second[n];
  map<long long, long long> mp;
  set<long long> st;
  for (int i = 0; i < n; i++) {
    cin >> second[i];
    mp[second[i]]++;
    st.insert(second[i]);
  }
  long long o = 0;
  long long e = 0;
  for (auto x : st) {
    if (mp[x] == 1) {
      o++;
    } else if (mp[x] > 2) {
      e++;
    }
  }
  bool first = true;
  if (o % 2 != 0) {
    if (e == 0) first = false;
  }
  if (first == false)
    cout << "NO\n";
  else {
    cout << "YES\n";
    char ans[n];
    for (int i = 0; i < n; i++) ans[i] = 'A';
    long long c = 0;
    for (int i = 0; i < n; i++) {
      if (mp[second[i]] == 1) {
        c++;
        if (c % 2 == 0) {
          ans[i] = 'B';
        }
      }
    }
    if (c % 2 != 0) {
      for (int i = 0; i < n; i++) {
        if (mp[second[i]] > 2) {
          ans[i] = 'B';
          break;
        }
      }
    }
    for (int i = 0; i < n; i++) cout << ans[i];
    cout << endl;
  }
  return 0;
}