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 n, a[109], cnt[109], temp, res;
bool used[109], mark[109];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (cnt[i] == 1) res++;
  }
  for (int i = 0; i < n && temp < res - 1; i++) {
    if (!used[a[i]] && cnt[a[i]] == 1) {
      temp++;
      used[a[i]] = true;
      res--;
      mark[i] = true;
    }
  }
  if (temp < res) {
    for (int i = 0; i < n; i++) {
      if (!used[a[i]] && cnt[a[i]] > 2) {
        temp++;
        used[a[i]] = true;
        mark[i] = true;
        break;
      }
    }
  }
  if (temp == res) {
    printf("YES\n");
    for (int j = 0; j < n; j++) {
      printf("%c", (mark[j] ? 'B' : 'A'));
    }
  } else
    cout << "NO";
  return 0;
}