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;
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) { return a + b - max(a, b); }
long long mod(long long a, long long b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  map<long long, long long> m1;
  map<long long, char> m2;
  long long n;
  cin >> n;
  long long x;
  set<long long> s2;
  long long ctr = 0, ctr1 = 0;
  set<long long> s1;
  vector<long long> vec;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    vec.push_back(x);
    s1.insert(x);
    m1[x]++;
  }
  long long pt1;
  for (auto it : s1) {
    if (m1[it] > 1) {
      m2[it] = 'A';
    } else {
      ctr++;
      if (ctr1 == 0) {
        m2[it] = 'A';
        ctr1 = 1;
      } else {
        m2[it] = 'B';
        ctr1 = 0;
      }
    }
  }
  bool flag = false;
  long long num = -1;
  for (auto it : s1) {
    if (m1[it] >= 3 && ctr % 2) {
      flag = true;
      num = it;
      break;
    }
  }
  if (ctr % 2 && flag == false) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (auto it : vec) {
    if (it == num && flag) {
      cout << 'B';
      flag = false;
    } else
      cout << m2[it];
  }
}