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;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char c) { return '\'' + string(1, c) + '\''; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) res += ", ";
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
string to_str(const int &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
string to_str(const long long &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
template <class T>
string tostring(T x, int len = 0) {
  stringstream ss;
  ss << x;
  string r = ss.str();
  if (int((r).size()) < len) r = string(len - int((r).size()), '0') + r;
  return r;
}
template <class T>
void convert(string x, T &r) {
  stringstream ss(x);
  ss >> r;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 998244353;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 998244353;
    a = a * a % 998244353;
  }
  return res;
}
long long modinv(long long a) { return powmod(a, 998244353 - 2); }
long long modinv(long long a, long long m) { return powmod(a, m - 2); }
long long nCkMp(long long n, long long k) {
  long long numerator = 1, denominator = 1;
  for (int i = 0; i < k; i++) numerator = (numerator * (n - i)) % 998244353;
  for (int i = 1; i <= k; i++) denominator = (denominator * i) % 998244353;
  return (numerator * modinv(denominator)) % 998244353;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
int TT = 1, tt;
clock_t t1;
void solve() {
  t1 = clock();
  int n, a = 0, c = 0;
  (cin >> (n));
  vector<int> x(n);
  for (auto i = (0); i < (n); i++) (cin >> (x[i]));
  vector<int> y(105);
  for (auto i = (0); i < (n); i++) y[x[i]]++;
  for (auto i = (0); i < (105); i++)
    if (y[i] == 1)
      a++;
    else if (y[i] > 2)
      c++;
  string z(n, 'A');
  if (a % 2 == 0) {
    a /= 2;
    (cout << ("YES") << endl);
    for (auto i = (0); i < (105); i++) {
      if (a == 0) break;
      if (y[i] == 1) {
        a--;
        for (auto j = (0); j < (n); j++)
          if (x[j] == i) z[j] = 'B';
      }
    }
    (cout << (z) << endl);
  } else if (a % 2 && c) {
    a /= 2;
    (cout << ("YES") << endl);
    for (auto i = (0); i < (105); i++) {
      if (a == 0) break;
      if (y[i] == 1) {
        a--;
        for (auto j = (0); j < (n); j++)
          if (x[j] == i) z[j] = 'B';
      }
    }
    404;
    for (auto i = (0); i < (105); i++) {
      if (y[i] > 2) {
        404;
        for (auto j = (0); j < (n); j++) {
          if (x[j] == i) {
            z[j] = 'B';
            break;
          }
        }
        break;
      }
    }
    404;
    (cout << (z) << endl);
  } else
    (cout << ("NO") << endl);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (tt = 0; tt < TT; tt++) solve();
  return EXIT_SUCCESS;
}