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;
int main() {
  int n;
  cin >> n;
  vector<int> v(n), w(100);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    w[v[i] - 1]++;
  }
  unordered_set<int> a, b, c, d;
  for (int i = 0; i < 100; ++i) {
    if (w[i] == 1) {
      a.insert(i);
      a.swap(b);
    } else if (w[i] == 2) {
      c.insert(i);
    } else if (w[i] > 0) {
      d.insert(i);
    }
  }
  if (a.size() != b.size() && d.empty()) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  int special = -1;
  if (a.size() < b.size()) {
    special = *d.begin();
  }
  a.insert(begin(c), end(c));
  for (int i = 0; i < n; ++i) {
    if (a.find(v[i] - 1) != a.end())
      cout << 'A';
    else if (v[i] - 1 == special) {
      cout << 'A';
      special = -1;
    } else
      cout << 'B';
  }
  cout << endl;
  return 0;
}
