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;
const int INF = 1e9 + 5;
const long long LINF = 1e18 + 5;
const double eps = 1e-9;
fstream fin, fout;
int main() {
  int n;
  cin >> n;
  vector<int> q(n);
  for (int &i : q) cin >> i;
  vector<int> one;
  vector<char> v(n, false);
  int d = -1;
  map<int, int> z;
  for (int i = 0; i < n; i++) z[q[i]]++;
  for (int i = 0; i < n; i++)
    if (z[q[i]] == 1)
      one.push_back(i);
    else if (z[q[i]] > 2)
      d = i;
  for (int i = 0; i < one.size() / 2; i++) v[one[i]] = 1;
  bool ans = 1;
  if (one.size() % 2 == 1) {
    if (d == -1)
      ans = 0;
    else
      v[d] = 1;
  }
  if (ans) {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) cout << char('A' + v[i]);
  } else
    cout << "NO";
}
