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 main() {
  int n;
  cin >> n;
  vector<pair<int, int> > q(n);
  for (int i = 0; i < n; i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  map<int, int> m;
  for (int i = 0; i < n; i++) m[q[i].first]++;
  int alrPerf = 0;
  int tw = 0;
  int g = 0;
  for (auto it : m) {
    if (it.second == 1) alrPerf++;
    if (it.second == 2) tw++;
    if (it.second > 2) g++;
  }
  int a = 0, b = 0;
  map<int, char> p;
  for (int i = 0; i < n; i++) {
    if (m[q[i].first] == 1)
      if (a >= b)
        p[q[i].first] = 'B', b++;
      else
        p[q[i].first] = 'A', a++;
  }
  bool twEx = false;
  bool fEx = false;
  if (a != b) {
    if (g) fEx = true;
  }
  if (a != b && !fEx) {
    cout << "NO";
    exit(0);
  }
  vector<char> alp(n, 'A');
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    if (m[q[i].first] == 1) alp[q[i].second] = p[q[i].first];
  }
  if (fEx) {
    for (int i = 0; i < n; i++)
      if (m[q[i].first] > 2) {
        if (a > b) {
          alp[q[i].second] = 'B';
          for (int j = i + 1; j < n && q[i].first == q[j].first; j++)
            alp[q[j].second] = 'A';
        } else {
          alp[q[i].second] = 'A';
          for (int j = i + 1; j < n && q[i].first == q[j].first; j++)
            alp[q[j].second] = 'B';
        }
        break;
      }
    for (int i = 0; i < n; i++) cout << alp[i];
    exit(0);
  }
  for (int i = 0; i < n; i++) cout << alp[i];
}