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() {
  int n, co = 0, ct = 0;
  cin >> n;
  int a[n], b[1001] = {0};
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[a[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (b[i] == 1) co++;
    if (b[i] >= 3) ct++;
  }
  if (!(co % 2)) {
    int t = co;
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if ((t > (co / 2)) && b[a[i]] == 1) {
        cout << "A";
        t--;
      } else if (b[a[i]] == 1)
        cout << "B";
      else
        cout << "A";
    }
    return 0;
  } else if (ct >= 1) {
    cout << "YES\n";
    int flag_final = 0, final = co + 1;
    co++;
    for (int i = 0; i < n; i++) {
      if (b[a[i]] == 1 && final > co / 2) {
        cout << "A";
        final--;
      } else if (b[a[i]] == 1)
        cout << "B";
      else if (b[a[i]] >= 3 && flag_final == 0) {
        flag_final = 1;
        cout << "B";
      } else
        cout << "A";
    }
    return 0;
  }
  cout << "NO\n";
  return 0;
}