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 local;
void solve() {
  int n, idx;
  cin >> n;
  int ar[n], counter[101];
  string s = "";
  memset(counter, 0, sizeof(counter));
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    counter[ar[i]]++;
    if (counter[ar[i]] > 2) {
      idx = i;
    }
  }
  int sum = 0;
  for (int i = 1; i <= 100; i++) {
    sum += (counter[i] == 1 || counter[i] == 2 ? counter[i] : 0);
  }
  if (local) cout << "sum = " << sum << endl;
  if (sum & 1 && n - sum == 0) {
    cout << "NO" << endl;
  } else {
    int side = 0;
    for (int i = 0; i < n; i++) {
      if (sum & 1 && n - sum != 0 && i == idx) {
        s += "B";
      } else if (counter[ar[i]] == 1) {
        if (side == 0)
          s += "A";
        else
          s += "B";
        side += 1;
        side %= 2;
      } else if (counter[ar[i]] == 2) {
        s += "A";
        counter[ar[i]] = -1;
      } else if (counter[ar[i]] == -1) {
        s += "B";
      } else
        s += "A";
    }
    cout << "YES" << endl;
    cout << s << endl;
  }
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  local = 0;
  if (local) {
    int tc;
    cin >> tc;
    while (tc--) solve();
  } else
    solve();
  return 0;
}