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 int N = 1e2 + 7;
int n;
int s[N];
int cnt[N];
int pos[N];
string ans;
int d;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  ans.resize(n);
  for (int i = 0; i < n; ++i) ans[i] = 'A';
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    if (cnt[s[i]] == 0) pos[s[i]] = i;
    ++cnt[s[i]];
  }
  for (int i = 0; i < N; ++i)
    if (cnt[i] == 1) ++d;
  for (int i = 0; i < N; ++i)
    if (cnt[i] == 1 && d >= 2) {
      ans[pos[i]] = 'B';
      d -= 2;
      --cnt[i];
    }
  for (int i = 0; i < N; ++i)
    if (cnt[i] >= 3 && d >= 1) {
      ans[pos[i]] = 'B';
      --d;
      --cnt[i];
    }
  if (d == 0)
    cout << "YES\n" << ans;
  else
    cout << "NO\n";
  return 0;
}
