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() {
  vector<int> vec[105];
  int arr[102];
  int fre[1002];
  int n, one = 0, more = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    arr[i] = x;
    vec[x].push_back(i);
  }
  for (int i = 0; i <= 100; i++) {
    if (vec[i].size() == 1) one++;
    if (vec[i].size() > 2) more++;
  }
  int now = 0;
  if (one % 2 == 1 && more == 0) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i <= 100; i++) {
    if (vec[i].size() == 1) {
      if (now == 0) {
        now = 1;
        fre[vec[i][0]] = 0;
      } else if (now == 1) {
        now = 0;
        fre[vec[i][0]] = 1;
      }
      vec[i].clear();
    }
  }
  if (one % 2 == 1) {
    for (int i = 0; i <= 100; i++) {
      if (vec[i].size() > 2) {
        if (now == 0) {
          fre[vec[i][0]] = 0;
          now = 1;
        } else if (now == 1) {
          fre[vec[i][0]] = 1;
          now = 0;
        }
        vec[i][0] = n + 1;
        break;
      }
    }
  }
  for (int i = 0; i <= 100; i++) {
    for (int j = 0; j < vec[i].size(); j++) fre[vec[i][j]] = now;
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << (char)(fre[i] + 'A');
}
