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;
vector<int> V;
int cont[101];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    V.push_back(tmp);
    cont[tmp]++;
  }
  int cont1 = 0;
  for (int i = 0; i < 101; i++) {
    if (cont[i] == 1) cont1++;
  }
  if (cont1 % 2 == 0) {
    cout << "YES\n";
    bool primo = true;
    for (int i = 0; i < n; i++) {
      if (cont[V[i]] == 1) {
        if (primo) {
          cout << "A";
          primo = false;
        } else {
          cout << "B";
          primo = true;
        }
      } else
        cout << "A";
    }
  } else {
    int index = -1;
    for (int i = 0; i < n; i++) {
      if (cont[V[i]] > 2) {
        index = i;
        break;
      }
    }
    if (index == -1) {
      cout << "NO";
      return 0;
    }
    cout << "YES\n";
    bool primo = true;
    for (int i = 0; i < n; i++) {
      if (cont[V[i]] == 1) {
        if (primo) {
          cout << "A";
          primo = false;
        } else {
          cout << "B";
          primo = true;
        }
      } else if (i == index)
        cout << "B";
      else
        cout << "A";
    }
  }
}