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() {
  int n = 0;
  cin >> n;
  vector<int> a(n);
  map<int, int> cc;
  for (int& aa : a) {
    cin >> aa;
    cc[aa]++;
  }
  int cnt = 0, cnt1 = 0;
  for (auto x : cc) {
    if (x.second == 1) {
      cnt++;
    } else if (x.second > 2) {
      cnt1++;
    }
  }
  if (cnt % 2 == 0) {
    cout << "YES" << endl;
    int x = 0;
    for (int i = 0; i < n; ++i) {
      if (cc[a[i]] == 1) {
        if (x % 2 == 0) {
          cout << 'A';
        } else {
          cout << 'B';
        }
        ++x;
      } else {
        cout << 'A';
      }
    }
  } else {
    if (cnt1 == 0) {
      cout << "NO" << endl;
      return 0;
    }
    cout << "YES" << endl;
    int x = 0;
    int val = 0;
    for (int i = 0; i < n; ++i) {
      if (cc[a[i]] == 1) {
        if (x % 2 == 0) {
          cout << 'A';
        } else {
          cout << 'B';
        }
        ++x;
      } else {
        if (cc[a[i]] > 2 && val == 0) {
          cout << 'B';
          val = 1;
        } else {
          cout << 'A';
        }
      }
    }
  }
}
