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;
  const int N = 101;
  int cnt[N] = {0};
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int cnt1 = 0, cnt2 = 0;
  for (int i = 0; i < N; i++) {
    cnt1 += cnt[i] == 1;
    cnt2 += cnt[i] == 2;
  }
  if (n - cnt1 - 2 * cnt2 == 0 && cnt1 % 2 == 1) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << '\n';
  if (cnt1 % 2 == 0) {
    int cntA = 0;
    for (int i = 0; i < n; i++)
      if (cnt[a[i]] == 2)
        cout << 'A';
      else if (cnt[a[i]] == 1 && 2 * cntA < cnt1) {
        cout << 'A';
        cntA++;
      } else
        cout << 'B';
  } else {
    char c[N] = {0};
    int cntA = 0;
    char f3 = -1, last;
    for (int i = 0; i < n; i++)
      if (cnt[a[i]] == 2)
        c[a[i]] = 'A';
      else if (cnt[a[i]] == 1 && cntA < (cnt1 + 1) / 2) {
        c[a[i]] = 'A';
        cntA++;
      } else {
        if (cnt[a[i]] == 1)
          c[a[i]] = 'B';
        else {
          if (c[a[i]] == 0) c[a[i]] = 'B';
          if (f3 == -1) {
            f3 = a[i];
            c[a[i]] = 'A';
          } else if (a[i] == f3)
            last = i;
        }
      }
    for (int i = 0; i < n; i++)
      if (i == last)
        cout << 'B';
      else
        cout << c[a[i]];
  }
}
