Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), f(101);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  int nc = 0;
  for (int i = 1; i <= 100; i++) {
    if (f[i] == 1) nc++;
  }
  if (nc % 2 == 0) {
    cout << "YES\n";
    int tag = 0;
    for (int i = 0; i < n; i++) {
      if (f[a[i]] > 1)
        cout << 'A';
      else {
        if (tag)
          cout << 'A';
        else
          cout << 'B';
        tag ^= 1;
      }
    }
    return 0;
  }
  int flag = 0;
  for (int i = 1; i <= 100; i++) {
    if (f[i] > 2) {
      flag = i;
      break;
    }
  }
  if (flag == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  f[flag]--;
  int tag = 0;
  char tmp;
  vector<char> ans(n, '#');
  for (int i = 0; i < n; i++) {
    if (f[a[i]] > 1 && a[i] != flag)
      continue;
    else {
      if (tag)
        ans[i] = 'A';
      else
        ans[i] = 'B';
      tag ^= 1;
      if (a[i] == flag) {
        tmp = ans[i];
        flag = -1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == '#') {
      if (tmp == 'A')
        ans[i] = 'B';
      else
        ans[i] = 'A';
    }
    cout << ans[i];
  }
  return 0;
}