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 a[123];
int sum[123];
int vis[123];
set<int> V1, V2, V3;
int n;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum[a[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (sum[i] == 1)
      V1.insert(i);
    else if (sum[i] == 2)
      V2.insert(i);
    else if (sum[i] >= 3)
      V3.insert(i);
  }
  if (V1.size() % 2 != 0 && V3.empty())
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    if (V1.size() % 2 == 0) {
      int ct = 0;
      for (int i = 1; i <= n; i++) {
        if (V1.count(a[i])) {
          if (ct < (int)V1.size() / 2)
            cout << "A";
          else
            cout << "B";
          ct++;
        } else
          cout << "A";
      }
    } else {
      int ct = 0;
      for (int i = 1; i <= n; i++) {
        if (V1.count(a[i])) {
          if (ct < (int)V1.size() / 2)
            cout << "A";
          else
            cout << "B";
          ct++;
        } else if (V2.count(a[i])) {
          if (vis[a[i]] == 0)
            cout << "A";
          else
            cout << "B";
          vis[a[i]]++;
        } else {
          if (a[i] == *V3.begin()) {
            if (vis[a[i]])
              cout << "B";
            else
              cout << "A";
            vis[a[i]]++;
          } else
            cout << "B";
        }
      }
    }
  }
  return 0;
}
