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;
vector<int> oc[109];
int wut[109];
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> s(n);
  int one, two, three;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    oc[s[i]].push_back(i);
  }
  one = two = three = 0;
  for (int i = 1; i <= 100; i++) {
    if (oc[i].size() == 1)
      one++;
    else if (oc[i].size() == 2)
      two++;
    else if (oc[i].size() > 2)
      three++;
  }
  bool flg = false;
  if (one % 2 == 1 && three == 0) {
    cout << "NO" << endl;
    return 0;
  }
  if (one % 2) flg = true;
  int parity = 0;
  for (int i = 1; i <= 100; i++) {
    if (oc[i].size() == 0) continue;
    if (oc[i].size() == 1) {
      wut[oc[i][0]] = parity;
      parity ^= 1;
    } else if (oc[i].size() == 2) {
      wut[oc[i][0]] = 0;
      wut[oc[i][1]] = 1;
    } else {
      for (int x : oc[i]) {
        wut[x] = parity ^ 1;
      }
      if (flg) {
        wut[oc[i][0]] = parity;
        parity ^= 1;
        flg = false;
      }
    }
  }
  assert(flg == 0);
  assert(parity == 0);
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    if (wut[i] == 0)
      cout << "A";
    else if (wut[i] == 1)
      cout << "B";
  }
  return 0;
}
