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;
  while (cin >> n) {
    vector<long long> v(n);
    map<long long, int> mp;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]]++;
    }
    int un = 0;
    for (auto x : mp) {
      if (x.second == 1) un++;
    }
    bool flag = false;
    if (un % 2 == 1) {
      for (auto x : mp) {
        if (x.second >= 3) {
          flag = true;
          break;
        }
      }
    }
    if ((un) % 2 == 1 && !flag) {
      cout << "NO\n";
    } else {
      if (flag) {
        un++;
      }
      string res = "";
      un /= 2;
      for (int i = 0; i < n; i++) {
        if (mp[v[i]] == 1) {
          if (un > 0) {
            res.push_back('A');
            un--;
          } else {
            res.push_back('B');
          }
        } else {
          if (mp[v[i]] >= 3 && flag) {
            res.push_back('B');
            flag = false;
          } else {
            res.push_back('A');
          }
        }
      }
      cout << "YES\n" << res << "\n";
    }
  }
}
