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 = 1e9 + 7;
long long const inf = 2e18;
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;
}
inline long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n & 1) {
    return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
  } else {
    return (modexp((x * x) % mod, n / 2) % mod);
  }
}
long long modinv(long long n) { return modexp(n, mod - 2); }
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  string second;
  map<long long, vector<long long> > m;
  long long mx = 0;
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
    second.push_back('0');
    m[a[i]].push_back(i);
    mx = max(mx, (long long)m[a[i]].size());
  }
  long long id = -1;
  for (long long i = 0; i < (n); i++) {
    if (m.find(a[i]) != m.end()) {
      if (m[a[i]].size() > 2) id = a[i];
      if (m[a[i]].size() == 2) {
        second[m[a[i]][0]] = 'A';
        second[m[a[i]][1]] = 'A';
        m.erase(a[i]);
      }
    }
  }
  long long x = 1;
  for (auto it : m) {
    if (id == it.first) continue;
    if (x) {
      second[it.second[0]] = 'A';
      for (int j = 1; j < it.second.size(); j++) {
        second[it.second[j]] = 'B';
      }
      if (it.second.size() == 2) continue;
    } else {
      second[it.second[0]] = 'B';
      for (int j = 1; j < it.second.size(); j++) {
        second[it.second[j]] = 'A';
      }
    }
    x = 1 ^ x;
  }
  if (x == 0) {
    if (id == -1) {
      cout << "NO";
      return;
    }
    second[m[id][0]] = 'B';
    for (int j = 1; j < m[id].size(); j++) {
      second[m[id][j]] = 'A';
    }
  } else {
    for (auto it : m[id]) {
      second[it] = 'A';
    }
  }
  cout << "YES\n" << second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}