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() {
  long long n, c = 0, f = 0;
  cin >> n;
  vector<long long> v;
  map<long long, long long> mp;
  map<long long, char> mpp;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    mp[p]++;
    v.push_back(p);
  }
  long long pp = -1, k = 0;
  for (auto i = mp.begin(); i != mp.end(); i++) {
    if (i->second > 2) {
      pp = i->first;
      k++;
    }
    if (i->second >= 2)
      mpp[i->first] = 'A';
    else
      c++;
  }
  if (c % 2 != 0) {
    if (!k) {
      cout << "NO" << endl;
      return 0;
    }
    for (auto i = mp.begin(); i != mp.end(); i++)
      if (i->second == 1) {
        if (!f)
          mpp[i->first] = 'A', f = 1;
        else
          mpp[i->first] = 'B', f = 0;
      }
  } else {
    for (auto i = mp.begin(); i != mp.end(); i++)
      if (i->second == 1) {
        if (!f)
          mpp[i->first] = 'A', f = 1;
        else
          mpp[i->first] = 'B', f = 0;
      }
  }
  long long cc = 0;
  cout << "YES" << endl;
  for (long long i = 0; i < n; i++) {
    if (v[i] == pp && cc == 0 && c % 2 != 0) {
      cc = 1;
      cout << "B";
      continue;
    } else
      cout << mpp[v[i]];
  }
}
