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;
  cin >> n;
  int arr[n], brr[101] = {0}, k = 0, crr[100] = {0}, drr[100] = {0}, k1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    brr[arr[i]]++;
  }
  for (int i = 0; i <= 100; i++) {
    if (brr[i] == 1) {
      crr[k] = i;
      k++;
    }
    if (brr[i] > 2) {
      drr[k1] = i;
      k1++;
    }
  }
  int kcopy = k;
  if (k % 2 == 0) {
    cout << "YES\n";
    int x = 0;
    for (int i = 0; i < n; i++) {
      int check = 0;
      if (brr[arr[i]] == 1) {
        x++;
        if (x <= kcopy / 2) {
          cout << "A";
        } else
          cout << "B";
      } else
        cout << "B";
    }
  } else if (k1 >= 1) {
    cout << "YES\n";
    int y = 0, stop = 0;
    for (int i = 0; i < n; i++) {
      if (brr[arr[i]] == 1) {
        y++;
        if (y <= kcopy / 2) {
          cout << "A";
        } else
          cout << "B";
      } else if (brr[arr[i]] > 2 && stop != 1) {
        cout << "A";
        stop = 1;
      } else
        cout << "B";
    }
  } else
    cout << "NO";
}