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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, temp = 0;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) m[a[i]]++;
  for (long long int i = 0; i < n; i++)
    if (m[a[i]] == 1) temp++;
  if (temp % 2) {
    long long int tt = 1, w = 0;
    for (long long int i = 0; i < n; i++) {
      if (m[a[i]] > 2) w = 1;
    }
    if (!w) {
      cout << "NO";
      return 0;
    }
    cout << "YES\n";
    w = 0;
    for (long long int i = 0; i < n; i++) {
      if (m[a[i]] > 2 && !w) {
        cout << 'B';
        w = 1;
      } else if (m[a[i]] > 1)
        cout << 'A';
      else if (m[a[i]] == 1 && tt <= temp / 2 + 1) {
        cout << 'A';
        tt++;
      } else
        cout << 'B';
    }
  } else {
    cout << "YES\n";
    long long int tt = 1, ch = 0;
    for (long long int i = 0; i < n; i++) {
      if (m[a[i]] > 1)
        cout << 'A';
      else if (m[a[i]] == 1 && tt <= temp / 2) {
        cout << 'A';
        tt++;
      } else {
        ch++;
        cout << 'B';
      }
    }
  }
}