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 q[1001], f, h;
int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    q[a[i]]++;
    if (q[a[i]] == 1)
      h++;
    else if (q[a[i]] == 2)
      h--;
    else if (q[a[i]] == 3 && f == 0)
      f = 1, a[i] = 1000;
  }
  if (h % 2 && f == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  string hh = "AB";
  int o = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == 1000 && h % 2)
      cout << "B";
    else if (q[a[i]] == 1) {
      cout << hh[o];
      o = (o + 1) % 2;
    } else
      cout << "A";
  }
}
