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;
  int ar[n];
  int visit[n];
  int a = 0, b = 0;
  memset(visit, -1, sizeof(visit));
  for (int i = 0; i < n; ++i) {
    cin >> ar[i];
  }
  bool f = true;
  string output(n, 'A');
  for (int i = 0; i < n; ++i) {
    int count = 0;
    for (int j = i; j < n; ++j) {
      if (ar[i] == ar[j] && visit[j] == -1) {
        count++;
        visit[j] = i;
      }
    }
    if (count == 1) {
      int first = 1;
      for (int j = i; j < n; ++j) {
        if (visit[j] == i) {
          if (f) {
            if (first)
              output[j] = 'A', a++;
            else
              output[j] = 'B';
          } else {
            if (first)
              output[j] = 'B', b++;
            else
              output[j] = 'A';
          }
          first = 0;
        }
      }
      f = !f;
    }
  }
  if (a == b)
    cout << "YES" << endl << output << endl;
  else {
    int need = abs(a - b);
    set<int> s;
    for (int i = 0; i < n && need; ++i) {
      int count = 0;
      if (s.count(ar[i])) continue;
      s.insert(ar[i]);
      for (int j = 0; j < n; ++j) {
        if (ar[i] == ar[j]) count++;
      }
      if (count > 2) {
        int f = 1;
        for (int j = 0; j < n; ++j) {
          if (ar[i] == ar[j]) {
            if (f == 1) {
              if (a < b) {
                output[j] = 'A';
              } else
                output[j] = 'B';
            } else {
              if (a < b) {
                output[j] = 'B';
              } else
                output[j] = 'A';
            }
            f = 0;
          }
        }
        need--;
      }
    }
    if (need == 0)
      cout << "YES" << endl << output << endl;
    else
      cout << "NO" << endl;
  }
}
