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;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> m;
  vector<bool> ans(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  long long pos = 0;
  long long z = 0;
  long long o = 0;
  for (long long i = 0; i < n; i++) {
    if (m[a[i]] == 1) {
      ans[i] = (pos & 1);
      pos++;
      z += (ans[i] == 0);
      o += (ans[i] == 1);
    } else
      ans[i] = 0;
  }
  if (z > o) {
    for (long long i = 0; i < n; i++) {
      if (m[a[i]] > 2) {
        ans[i] = 1;
        o++;
        break;
      }
    }
  } else if (o > z) {
    for (long long i = 0; i < n; i++) {
      if (m[a[i]] > 2) {
        ans[i] = 0;
        z++;
        break;
      }
    }
  }
  if (o == z) {
    cout << "YES"
         << "\n";
    for (long long i = 0; i < n; i++) cout << char(ans[i] + 'A');
  } else
    cout << "NO"
         << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}