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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  vector<int> a;
  vector<int> b(210, 0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    a.push_back(temp);
    b[temp]++;
  }
  int count1 = 0, count = 0;
  for (int i = 0; i < b.size(); i++) {
    if (b[i] == 1)
      count1++;
    else if (b[i] > 2)
      count++;
  }
  if (count1 % 2 != 0 && count == 0)
    cout << "NO" << endl;
  else {
    vector<char> re;
    if (count1 % 2 != 0) {
      int t1 = 0, t2 = 0;
      for (int i = 0; i < n; i++) {
        if (b[a[i]] == 1 && t1 < count1 / 2) {
          re.push_back('B');
          t1++;
        } else if (b[a[i]] > 2 && t2 == 0) {
          re.push_back('B');
          t2++;
        } else {
          re.push_back('A');
        }
      }
    } else if (count1 % 2 == 0) {
      int t = 0;
      for (int i = 0; i < n; i++) {
        if (b[a[i]] == 1 && t < count1 / 2) {
          re.push_back('B');
          t++;
        } else
          re.push_back('A');
      }
    }
    cout << "YES" << endl;
    for (int i = 0; i < re.size(); i++) cout << re[i];
    cout << endl;
  }
  return 0;
}