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;
const int N = 109;
int n;
int a[N];
map<int, int> m;
set<int> s[2];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    m[a[i]]++;
  }
  int pos = 0;
  for (auto p : m)
    if (p.second == 1) {
      s[pos].insert(p.first);
      pos = 1 - pos;
    }
  if (s[0].size() == s[1].size()) {
    string res(n, 'A');
    for (int i = 0; i < n; ++i)
      if (s[1].count(a[i])) res[i] = 'B';
    cout << "YES" << endl;
    for (auto c : res) cout << c;
    cout << endl;
    return 0;
  } else {
    assert(int(s[0].size()) - 1 == int(s[1].size()));
    string res(n, 'A');
    for (int i = 0; i < n; ++i)
      if (s[1].count(a[i])) res[i] = 'B';
    int id = -1;
    for (auto p : m)
      if (p.second >= 3) {
        id = p.first;
        break;
      }
    if (id == -1) {
      cout << "NO" << endl;
      return 0;
    }
    bool flag = false;
    for (int i = 0; i < n; ++i) {
      if (a[i] == id)
        if (!flag) {
          flag = true;
          res[i] = 'B';
        }
    }
    cout << "YES" << endl;
    for (auto c : res) cout << c;
    cout << endl;
    return 0;
  }
  return 0;
}