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
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, i;
  cin >> n;
  vector<long long int> v[105];
  char c[n];
  for (i = 0; i < n; i++) c[i] = 'A';
  for (i = 0; i < n; i++) {
    long long int y;
    cin >> y;
    v[y].push_back(i);
  }
  long long int single = 0, odd = 0, even = 0;
  for (i = 1; i <= 100; i++) {
    if ((long long int)v[i].size() == 1) single++;
    if ((long long int)v[i].size() % 2 != 0 && (long long int)v[i].size() > 1)
      odd++;
    if ((long long int)v[i].size() % 2 == 0 && (long long int)v[i].size() > 2)
      even++;
  }
  if (single % 2 == 0) {
    single /= 2;
    for (i = 1; i <= 100 && single > 0; i++) {
      if ((long long int)v[i].size() == 1) c[v[i][0]] = 'B', single--;
    }
  } else {
    if (odd == 0 && even == 0) {
      cout << "NO";
      return 0;
    }
    single /= 2;
    for (i = 1; i <= 100 && single > 0; i++) {
      if ((long long int)v[i].size() == 1) c[v[i][0]] = 'B', single--;
    }
    if (odd > 0) {
      for (i = 1; i <= 100; i++) {
        if ((long long int)v[i].size() % 2 != 0 &&
            (long long int)v[i].size() > 1) {
          c[v[i][0]] = 'B';
          break;
        }
      }
    } else {
      for (i = 1; i <= 100; i++) {
        if ((long long int)v[i].size() % 2 == 0 &&
            (long long int)v[i].size() > 2) {
          c[v[i][0]] = 'B';
          break;
        }
      }
    }
  }
  cout << "YES\n";
  for (i = 0; i < n; i++) cout << c[i];
}
