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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> s(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
  }
  map<int, int> mp;
  set<int> a;
  set<int> b;
  for (int i = 1; i <= n; ++i) {
    ++mp[s[i]];
  }
  bool p = true;
  for (pair<int, int> z : mp) {
    int x = z.first;
    int y = z.second;
    if (y == 1) {
      if (p) {
        a.insert(x);
      } else {
        b.insert(x);
      }
      p = !p;
    }
  }
  if (a.size() == b.size()) {
    cout << "YES\n";
    for (int i = 1; i <= n; ++i) {
      if (a.count(s[i]) == 1) {
        cout << "A";
      } else {
        cout << "B";
      }
    }
    cout << "\n";
  } else {
    int p = 0;
    for (pair<int, int> z : mp) {
      int x = z.first;
      int y = z.second;
      if (y > 2) {
        p = x;
        break;
      }
    }
    if (p == 0) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      bool q = false;
      for (int i = 1; i <= n; ++i) {
        if (a.count(s[i]) == 1) {
          cout << "A";
        } else if (s[i] == p) {
          if (q == false) {
            cout << "B";
            q = true;
          } else {
            cout << "A";
          }
        } else {
          cout << "B";
        }
      }
      cout << "\n";
    }
  }
  return 0;
}
