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>
int32_t main() {
  int n;
  scanf("%d", &n);
  std::map<int, std::vector<int> > log;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    log[a].push_back(i);
  }
  std::vector<bool> ans(n);
  int cnt0 = 0, cnt1 = 0;
  for (auto p : log) {
    if (p.second.size() != 1) continue;
    if (cnt0 < cnt1) {
      cnt0++;
      p.second.pop_back();
    } else {
      cnt1++;
      ans[p.second.back()] = 1;
      p.second.pop_back();
    }
  }
  for (auto p : log) {
    if (p.second.size() <= 2) continue;
    if (cnt0 < cnt1) {
      cnt0++;
      p.second.pop_back();
      while (!p.second.empty()) {
        ans[p.second.back()] = 1;
        p.second.pop_back();
      }
    } else if (cnt0 > cnt1) {
      ans[p.second.back()] = 1;
      p.second.pop_back();
      cnt1++;
      while (!p.second.empty()) p.second.pop_back();
    }
  }
  if (cnt0 != cnt1) return puts("NO"), 0;
  puts("YES");
  for (bool b : ans) putchar(b ? 'A' : 'B');
  return puts(""), 0;
}