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, a, b;
  cin >> n;
  int arr[n];
  int z[101];
  for (int i = 0; i < 101; i++) z[i] = 0;
  b = n;
  int k = 0;
  while (b--) {
    cin >> a;
    arr[k] = a;
    ++k;
    ++z[a];
  }
  pair<int, int> s[101];
  for (int i = 0; i < 101; i++) s[i] = make_pair(z[i], i);
  sort(s, s + 101);
  int count = 0;
  for (int i = 0; i < 101; i++) {
    if (s[i].first == 1) ++count;
    if (s[i].first > 1) break;
  }
  unordered_map<int, int> A;
  int counter = 0;
  for (int i = 0; i < 101; i++) {
    if (counter < count / 2 and s[i].first == 1) {
      ++counter;
      ++A[s[i].second];
    }
    if (counter == count / 2) break;
  }
  bool flag = false;
  if (count % 2 == 0) flag = true;
  if (count % 2 == 1) {
    for (int i = 0; i < 101; i++)
      if (s[i].first >= 3) {
        ++A[s[i].second];
        flag = true;
        break;
      }
  }
  if (flag) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if (A[arr[i]] > 0) {
        cout << "A";
        --A[arr[i]];
      } else
        cout << "B";
    }
  } else
    cout << "NO\n";
}
