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;
int main() {
  long long n;
  std::cin >> n;
  vector<long long> sa(n);
  map<long long, long long> cou;
  for (int i = 0; i < n; i++) {
    std::cin >> sa[i];
    cou[sa[i]]++;
  }
  long long a = 0;
  long long c = 0;
  for (auto e : cou) {
    if (e.second >= 3) {
      c++;
    } else if (e.second == 1) {
      a++;
    }
  }
  long long ai, ak;
  bool ok = false;
  for (int i = 0; i <= a; i++) {
    for (int k = -c; k <= c; k++) {
      if (i + k == a - i) {
        ai = i;
        ak = k;
        ok = true;
        break;
      }
    }
  }
  if (!ok) {
    std::cout << "NO" << std::endl;
    return 0;
  }
  std::cout << "YES" << std::endl;
  map<long long, bool> done;
  for (int i = 0; i < n; i++) {
    if (cou[sa[i]] == 1) {
      if (ai > 0) {
        std::cout << 'A';
        ai--;
      } else {
        std::cout << 'B';
      }
    } else if (cou[sa[i]] == 2) {
      std::cout << 'A';
    } else {
      if (ak > 0 && done[sa[i]]) {
        std::cout << 'B';
        continue;
      }
      if (ak > 0 && !done[sa[i]]) {
        std::cout << 'A';
        ak--;
        done[sa[i]] = true;
        continue;
      }
      if (ak < 0 && done[sa[i]]) {
        std::cout << 'A';
        continue;
      }
      if (ak < 0 && !done[sa[i]]) {
        std::cout << 'B';
        ak++;
        done[sa[i]] = true;
        continue;
      }
      if (ak == 0) {
        std::cout << 'A';
      }
    }
  }
  std::cout << std::endl;
}