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;
const long long int maxn = 2e2 + 10;
const long long int modn = 1e9 + 7;
long long int T, len, num, x, sum, fir, sec;
long long int perm[maxn];
long long int all[maxn];
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> len;
  for (int i = 1; i <= len; i++) {
    cin >> perm[i];
    all[perm[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (all[i] == 1)
      fir++;
    else if (all[i] > 2)
      sec++;
  }
  if (fir % 2 == 0) {
    cout << "YES" << endl;
    for (int i = 1; i <= len; i++) {
      if (all[perm[i]] == 1 and fir % 2 == 1) {
        cout << "A";
        fir--;
      } else if (all[perm[i]] == 1 and fir % 2 == 0) {
        cout << "B";
        fir--;
      } else
        cout << "A";
    }
    return 0;
  }
  if (fir % 2 == 1 and sec > 0) {
    cout << "YES" << endl;
    for (int i = 1; i <= len; i++) {
      if (all[perm[i]] > 2 and sec > 0) {
        cout << "B";
        sec = 0;
      } else if (all[perm[i]] == 1 and fir % 2 == 1) {
        cout << "A";
        fir--;
      } else if (all[perm[i]] == 1 and fir % 2 == 0) {
        cout << "B";
        fir--;
      } else
        cout << "A";
    }
    return 0;
  }
  cout << "NO";
  return 0;
}
