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;
long long zz;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, n;
  cin >> n;
  char s[n + 1];
  for (i = 0; i < n; i++) s[i] = 'A';
  s[n] = '\0';
  int arr[n];
  int count[101] = {0};
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    count[arr[i]]++;
  }
  bool visited[101] = {0};
  vector<int> single;
  vector<int> dubs;
  for (i = 0; i < n; i++) {
    if (count[arr[i]] == 1) {
      single.push_back(i);
    } else if (count[arr[i]] > 2 && visited[arr[i]] == false) {
      dubs.push_back(i);
      visited[arr[i]] = true;
    }
  }
  if (single.size() % 2 == 0) {
    cout << "YES"
         << "\n";
    for (i = 0; i < (single.size()) / 2; i++) {
      s[single[i]] = 'B';
    }
    cout << s << "\n";
  } else if (!dubs.empty()) {
    cout << "YES"
         << "\n";
    for (i = 0; i < (single.size()) / 2; i++) {
      s[single[i]] = 'B';
    }
    s[dubs[0]] = 'B';
    cout << s << "\n";
  } else {
    cout << "NO"
         << "\n";
  }
  return 0;
}