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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, cnt1 = 0, cnt3 = 0, cnt2 = 0;
  cin >> n;
  vector<int> arr(n);
  vector<char> ans(n, 'B');
  map<int, int> a;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    a[arr[i]]++;
  }
  for (map<int, int>::iterator it = a.begin(); it != a.end(); ++it) {
    int f = it->first;
    int g = it->second;
    if (g == 1) cnt1++;
    if (g > 2) cnt3++;
  }
  if (cnt1 % 2 and cnt3 == 0) {
    return cout << "NO", 0;
  }
  cout << "YES\n";
  if (cnt1 % 2) {
    bool is = 0;
    int l = 0;
    for (int i = 0; i < n; i++) {
      if (a[arr[i]] > 2 and !is) {
        is = 1;
        ans[i] = 'A';
      }
      if (a[arr[i]] == 1 and l < cnt1 - 1) {
        ans[i] = 'A';
        cnt1--;
        l++;
      }
      cout << ans[i];
    }
    return 0;
  }
  int l = 0;
  for (int i = 0; i < n; i++) {
    if (a[arr[i]] == 1 && l < cnt1) {
      ans[i] = 'A';
      cnt1--;
      l++;
    }
    cout << ans[i];
  }
  return 0;
}
