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;
using ll = long long;
using ld = long double;
template <class T>
bool remin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool remax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
std::mt19937 rng(
    (long long)std::chrono::steady_clock::now().time_since_epoch().count());
const long long N = 2e5 + 2;
const ld eps = 1e-8;
void solve_case() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> f[103];
  vector<bool> c(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    f[a[i]].push_back(i);
  }
  long long ct = 0, c0 = 0, c1 = 0;
  for (long long i = 0; i < n; i++) {
    if (f[a[i]].size() == 1) {
      c[i] = (ct & 1);
      ct++;
      c0 += (c[i] == 0);
      c1 += (c[i] == 1);
    } else {
      c[i] = 0;
    }
  }
  if (c0 > c1) {
    for (long long i = 0; i < n; i++) {
      if (f[a[i]].size() > 2) {
        c[i] = 1;
        c1++;
        break;
      }
    }
  } else if (c1 > c0) {
    for (long long i = 0; i < n; i++) {
      if (f[a[i]].size() > 2) {
        c[i] = 0;
        c0++;
        break;
      }
    }
  }
  if (c0 != c1) return void(cout << "NO\n");
  cout << "YES\n";
  for (long long i = 0; i < n; i++) cout << char('A' + c[i]);
  cout << '\n';
}
signed main() {
  std::ios::sync_with_stdio(0);
  std::cout.tie(0);
  std::cin.tie(0);
  long long Testcases = 1;
  while (Testcases--) {
    solve_case();
  }
  return 0;
}