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 main() {
  int n;
  cin >> n;
  int a[n + 1];
  int cnt[1000] = {0};
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  vector<int> odd;
  int lit = -1;
  int ans[1000] = {0};
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]] == 1) {
      odd.push_back(a[i]);
    }
    if (cnt[a[i]] > 2) lit = i;
  }
  if (odd.size() % 2 != 0 and lit == -1) {
    cout << "NO\n";
    return 0;
  } else {
    string res = "";
    if (odd.size() % 2 == 0) {
      bool aa = false, bb = false;
      for (auto i : odd) {
        if (!aa) {
          ans[i] = 1;
          aa = true;
        } else {
          aa = false;
        }
      }
      for (int i = 1; i <= n; i++) {
        if (ans[a[i]] == 1)
          res += 'A';
        else
          res += 'B';
      }
    } else {
      bool aa = false, bb = false;
      for (auto i : odd) {
        if (!aa) {
          ans[i] = 1;
          aa = true;
        } else {
          aa = false;
        }
      }
      bool li = false;
      for (int i = 1; i <= n; i++) {
        if (ans[a[i]] == 1 or a[i] == a[lit])
          res += 'A';
        else {
          res += 'B';
        }
      }
      res[lit - 1] = 'B';
    }
    cout << "YES\n";
    cout << res << "\n";
  }
  return 0;
}
