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;
const int OO = 0x3f3f3f3f, NegOO = -1 * OO, N = 1e3 + 5, mod = 1e9 + 7;
int c[110], a[110], t[110];
int main() {
  ios_base::sync_with_stdio(0), ios::sync_with_stdio(0), cin.tie(0),
      cout.tie(0);
  int n;
  cin >> n;
  int p = -1;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> a[i];
    x = a[i];
    c[x]++;
    if (c[x] >= 3) p = x;
  }
  if ((n % 2 != 0) && p == -1) {
    return cout << "NO" << '\n', 0;
  }
  int f = 1;
  for (int i = 1; i <= 100; i++) {
    if (c[i] == 1) {
      if (f == 1) t[i] = 1;
      f = -f;
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 0; i < n; i++) {
    if (a[i] == p) {
      if (f == -1)
        cout << 'A';
      else
        cout << 'B';
      f = 1;
      continue;
    }
    if (t[a[i]])
      cout << 'B';
    else
      cout << 'A';
    t[a[i]] = 0;
  }
  return 0;
}
