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 n;
int count_input[101] = {0};
int input[101] = {0};
int countOnce = 0, countTwice = 0;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> input[i];
    count_input[input[i]]++;
  }
  for (int i = 1; i <= 101; i++) {
    if (count_input[i] == 1)
      countOnce++;
    else if (count_input[i] == 2)
      countTwice++;
  }
  if (countOnce % 2 == 0) {
    cout << "YES" << endl;
    int A = 0;
    for (int i = 1; i <= n; i++) {
      if (count_input[input[i]] == 1) {
        if (A < countOnce / 2) {
          cout << 'A';
          A++;
        } else
          cout << 'B';
      } else {
        cout << 'A';
      }
    }
    cout << endl;
  } else {
    if (countOnce + countTwice * 2 == n) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      int A = 0;
      bool bA = true;
      for (int i = 1; i <= n; i++) {
        if (count_input[input[i]] == 1) {
          if (A < countOnce / 2) {
            cout << 'A';
            A++;
          } else
            cout << 'B';
        } else if (count_input[input[i]] == 2) {
          cout << 'A';
        } else {
          if (bA) {
            cout << 'A';
            bA = false;
          } else
            cout << 'B';
        }
      }
      cout << endl;
    }
  }
  return 0;
}
