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>
#pragma GCC optimize("O3")
using namespace std;
int main() {
  map<int, int> mp;
  bool vis = false;
  int c1 = 0, y, c2 = 0, o = 0, n, t = 0, a[103];
  char x;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], mp[a[i]]++;
  for (int i = 0; i < n; i++)
    if (mp[a[i]] == 1)
      o++;
    else if (mp[a[i]] > 2)
      t++;
  if (o % 2) {
    if (t == 0) return cout << "NO", 0;
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if (mp[a[i]] == 1) {
        if (c1 < c2)
          cout << 'A', c1++;
        else
          cout << 'B', c2++;
      } else if (mp[a[i]] == 2)
        cout << 'A';
      else {
        if (vis) {
          if (a[i] == y)
            cout << x;
          else
            cout << 'A';
        } else {
          y = a[i];
          if (c1 < c2)
            cout << 'A', c1++, x = 'B';
          else
            cout << 'B', c2++, x = 'A';
          vis = true;
        }
      }
    }
  } else {
    cout << "YES\n";
    for (int i = 0; i < n; i++)
      if (mp[a[i]] > 1)
        cout << 'A';
      else if (c1 < c2)
        cout << 'A', c1++;
      else
        cout << 'B', c2++;
  }
  return 0;
}