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;
string place[2] = {"A", "B"};
int main() {
  long long int n;
  cin >> n;
  vector<long long int> ar(n);
  for (long long int i = 0; i < n; i++) cin >> ar[i];
  map<long long int, long long int> cnt;
  for (auto &it : ar) cnt[it]++;
  long long int sp = 0, imb = 0;
  map<long long int, long long int> SP, IMB;
  for (auto &it : cnt) {
    if (it.second == 1) {
      sp++;
      SP[it.first];
    } else if (it.second >= 3) {
      imb++;
      IMB[it.first];
    }
  }
  if (sp % 2 == 0) {
    string s = "";
    for (auto &it : ar) {
      if (SP.find(it) != SP.end()) {
        s = s + place[sp % 2];
        sp--;
      } else {
        s = s + "A";
      }
    }
    cout << "YES\n" << s;
  } else {
    if (imb == 0) {
      cout << "NO";
    } else {
      string s = "";
      for (auto &it : ar) {
        if (SP.find(it) != SP.end()) {
          s = s + place[sp % 2];
          sp--;
        } else {
          s = s + "B";
        }
      }
      for (long long int i = 0; i < n; i++) {
        if (IMB.find(ar[i]) != IMB.end()) {
          s[i] = 'A';
          break;
        }
      }
      cout << "YES\n" << s;
    }
  }
  return 0;
}