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;
long long a[200];
long long b[200];
long long c[200];
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]]++;
  }
  long long cnt = 0;
  long long e = 0;
  for (long long i = 1; i <= 100; i++) {
    if (b[i] == 1) cnt++;
    if (b[i] > 2) e = 1;
  }
  if (cnt % 2 == 0) {
    long long r = cnt / 2;
    cout << "YES" << endl;
    for (long long i = 1; i <= n; i++) {
      if (b[a[i]] != 1 || r == 0)
        cout << "A";
      else {
        cout << "B";
        r--;
      }
    }
  } else {
    if (!e)
      cout << "NO" << endl;
    else {
      long long r = cnt / 2;
      long long rr = 1;
      cout << "YES" << endl;
      for (long long i = 1; i <= n; i++) {
        if (b[a[i]] == 1 && r) {
          cout << "A";
          r--;
        } else {
          if (b[a[i]] >= 3 && rr) {
            cout << "A";
            rr--;
          } else
            cout << "B";
        }
      }
    }
  }
  return 0;
}