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;
const long double pi = acos((long double)-1.0);
const double eps = (double)1e-9;
const int inf = (int)1e9 + 7;
const int fastio = []() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return 0;
}();
int main(void) {
  int n, k = 0;
  cin >> n;
  int cnt[101] = {
      0,
  };
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int ans = 0;
  for (int i = 0; i < 101; i++) {
    if (cnt[i] == 1) k++;
    if (cnt[i] > 2) ans = 1;
  }
  if (k % 2 == 0) {
    cout << "YES" << '\n';
    int c = 0;
    for (int i = 0; i < n; i++) {
      if (cnt[a[i]] == 1) {
        if (c >= k / 2) {
          cout << "B";
        } else
          cout << "A";
        c++;
      } else
        cout << "A";
    }
  } else if (ans) {
    cout << "YES" << '\n';
    int c = 0;
    int res = 1;
    for (int i = 0; i < n; i++) {
      if (cnt[a[i]] == 1) {
        if (c > k / 2) {
          cout << "B";
        } else
          cout << "A";
        c++;
      } else if (cnt[a[i]] > 2)
        cout << (res ? "B" : "A"), res = 0;
      else
        cout << "A";
    }
  } else
    cout << "NO";
  return 0;
}
