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;
  cin >> n;
  int a[n], count[100];
  bool flag = true;
  for (int i = 0; i < 100; i++) count[i] = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    count[a[i] - 1]++;
  }
  int sum1 = 0, sum2 = 0;
  for (int i = 0; i < 100; i++) {
    if (count[i] == 1)
      sum1++;
    else if (count[i] > 2)
      sum2++;
  }
  if (sum1 % 2 == 0 || sum2 > 0) {
    cout << "YES" << endl;
    int j = 0, lim = sum1 / 2;
    if (sum1 % 2 == 0) {
      for (int i = 0; i < n; i++) {
        if (count[a[i] - 1] == 1 && j < lim) {
          cout << "B";
          j++;
        } else
          cout << "A";
      }
    } else {
      for (int i = 0; i < n; i++) {
        if (count[a[i] - 1] == 1 && j < lim) {
          cout << "B";
          j++;
        } else if (count[a[i] - 1] > 2 && flag) {
          cout << "B";
          flag = false;
        } else
          cout << "A";
      }
    }
  } else
    cout << "NO";
  return 0;
}
