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;
int ct[101];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  int s[n];
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    ct[s[i]]++;
  }
  int ct1 = 0;
  for (int i = 0; i < 101; i++) {
    if (ct[i] == 1) ct1++;
  }
  bool f = 0;
  int c1 = 0, c2 = 0;
  string ans = "";
  if (ct1 % 2 == 0) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if (ct[s[i]] == 1) {
        if (f) {
          ans += 'A';
          f = 1 - f;
        } else {
          f = 1 - f;
          ans += 'B';
        }
      } else {
        ans += 'A';
      }
    }
    cout << ans;
  } else {
    bool f = 0;
    for (int i = 0; i < n; i++) {
      if (ct[s[i]] == 1) {
        if (f) {
          ans += 'A';
          f = 1 - f;
          c1++;
        } else {
          f = 1 - f;
          ans += 'B';
          c2++;
        }
      } else {
        ans += 'A';
      }
    }
    for (int i = 0; i < n; i++) {
      if (ct[s[i]] > 2) {
        for (int j = i + 1; j < n; j++) {
          if (s[j] == s[i]) ans[j] = 'B';
        }
        c1++;
        break;
      }
    }
    if (c1 == c2) {
      cout << "YES\n";
      cout << ans;
    } else
      cout << "NO";
  }
  return 0;
}