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;
void assign(vector<int>& v, char c, vector<char>& ans) {
  for (int i : v) ans[i] = c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  map<int, vector<int>> f;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int ai;
    cin >> ai;
    f[ai].push_back(i);
  }
  vector<char> ans(n);
  vector<vector<int>> ones, twos, threes;
  for (auto& p : f) {
    if (p.second.size() == 1)
      ones.push_back(p.second);
    else if (p.second.size() == 2)
      twos.push_back(p.second);
    else
      threes.push_back(p.second);
  }
  int a = 0, b = 0;
  for (vector<int>& v : ones) {
    if (a < b) {
      assign(v, 'A', ans);
      a++;
    } else {
      assign(v, 'B', ans);
      b++;
    }
  }
  for (vector<int>& v : twos) {
    assign(v, 'A', ans);
  }
  for (vector<int>& v : threes) {
    if (a < b) {
      ans[v.back()] = 'A';
      v.pop_back();
      assign(v, 'B', ans);
      a++;
    } else {
      assign(v, 'A', ans);
    }
  }
  if (a == b) {
    cout << "YES\n";
    for (char c : ans) cout << c;
    cout << '\n';
  } else {
    cout << "NO\n";
  }
  return 0;
}