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;
const int INF = (int)1e9 + 1;
const int MAXN = (int)1e5;
int al[101];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int ns[n];
  for (int i = (0); i < (int)(n); i++) cin >> ns[i], al[ns[i]]++;
  int mlt = -1, k = 0;
  for (int i = (0); i < (int)(n); i++) {
    if (al[ns[i]] > 2) mlt = ns[i];
    if (al[ns[i]] == 1) k++;
  }
  if (k & 1 && mlt < 0) cout << "NO", exit(0);
  cout << "YES" << endl;
  int sum = 0, f = 1;
  if (k & 1) {
    for (int i = (0); i < (int)(n); i++) {
      if (al[ns[i]] == 1) {
        if (sum < k / 2)
          cout << 'A';
        else
          cout << 'B';
        sum++;
      } else {
        if (ns[i] == mlt && f)
          cout << 'A', f = 0;
        else
          cout << 'B';
      }
    }
  } else {
    for (int i = (0); i < (int)(n); i++) {
      if (al[ns[i]] == 1) {
        if (sum < k / 2)
          cout << 'A';
        else
          cout << 'B';
        sum++;
      } else {
        cout << 'A';
      }
    }
  }
}