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 a, s, d[111], f[111], g[111], h, j, k, l, i, n, m;
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a;
    f[i] = a;
    d[a]++;
  }
  for (i = 1; i <= 100; i++) {
    if (d[i] == 1) k++;
    if (d[i] > 2) l++;
  }
  if (k % 2 && !l) {
    cout << "NO";
    return 0;
  }
  if (k % 2) j = 1;
  cout << "YES\n";
  for (i = 0; i < n; i++) {
    if (d[f[i]] == 1) {
      if (k % 2 == 1)
        cout << "A";
      else
        cout << "B";
      k--;
    } else if (d[f[i]] > 2 && j) {
      j = 0;
      cout << "B";
    } else
      cout << "A";
  }
}