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;
map<long long int, long long int> m1, m2;
vector<long long int> v;
long long int n, x, pos, fin, idx = -1;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    m1[x]++;
    v.push_back(x);
  }
  for (auto it = m1.begin(); it != m1.end(); it++) {
    if (it->second == 1) pos++;
  }
  if (pos % 2 != 0) {
    for (int i = 0; i < v.size(); i++) {
      if (m1[v[i]] > 2) {
        idx = v[i];
        break;
      }
    }
    if (idx == -1)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      fin = pos / 2;
      for (int i = 0; i < v.size(); i++) {
        if (v[i] == idx) {
          if (m1[v[i]] == 1)
            cout << "B";
          else
            cout << "A";
          m1[v[i]]--;
        } else if (m1[v[i]] == 1) {
          if (pos > fin)
            cout << "A";
          else
            cout << "B";
          pos--;
        } else {
          cout << "A";
        }
      }
    }
  } else {
    cout << "YES" << endl;
    fin = pos / 2;
    for (int i = 0; i < v.size(); i++) {
      if (m1[v[i]] == 1) {
        if (pos > fin)
          cout << "A";
        else
          cout << "B";
        pos--;
      } else
        cout << "A";
    }
  }
  return 0;
}
