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>
void input(std::size_t &n, std::vector<int> &nums) {
  std::cin >> n;
  nums.resize(n);
  for (size_t i = 0; i < n; i++) {
    std::cin >> nums[i];
  }
}
void output(std::vector<char> &res) {
  if (res.empty()) {
    std::cout << "NO" << std::endl;
  } else {
    std::cout << "YES" << std::endl;
    for (auto r : res) {
      std::cout << r;
    }
    std::cout << std::endl;
  }
}
std::vector<char> solve(std::size_t n, std::vector<int> &nums) {
  std::unordered_map<int, int> counts;
  for (auto n : nums) {
    counts[n]++;
  }
  int uniques = 0, many = -1;
  for (auto p : counts) {
    if (p.second == 1)
      uniques++;
    else if (p.second >= 3)
      many = p.first;
  }
  std::vector<char> res;
  res.resize(n);
  if (uniques % 2 == 0) {
    int acnt = 0;
    for (size_t i = 0; i < n; i++) {
      if (counts[nums[i]] == 1 && acnt < uniques / 2) {
        res[i] = 'A';
        acnt++;
      } else if (counts[nums[i]] == 1 && acnt == uniques / 2) {
        res[i] = 'B';
      } else {
        res[i] = 'A';
      }
    }
    return res;
  } else if (many != -1) {
    int acnt = 0;
    for (size_t i = 0; i < n; i++) {
      if (counts[nums[i]] == 1 && acnt < uniques / 2) {
        res[i] = 'A';
        acnt++;
      } else if (counts[nums[i]] == 1) {
        res[i] = 'B';
      } else if (nums[i] == many) {
        res[i] = 'A';
        many = -1;
      } else {
        res[i] = 'B';
      }
    }
    return res;
  } else {
    res.resize(0);
    return res;
  }
}
int main() {
  std::size_t n = 0;
  std::vector<int> nums;
  input(n, nums);
  std::vector<char> res = solve(n, nums);
  output(res);
  return 0;
}