Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int cnt[102];
int a[102];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i <= 100; i++) cnt[i] = 0;
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int good = 0;
  vector<int> v1, v2;
  for (int i = 0; i <= 100; i++)
    if (cnt[i] == 1) {
      good++;
      v1.push_back(i);
    }
  int semigood = 0;
  for (int i = 0; i <= 100; i++)
    if (cnt[i] > 2) {
      semigood++;
      v2.push_back(i);
    }
  bool ok[102];
  for (int i = 0; i <= 100; i++) ok[i] = false;
  if (!(good & 1)) {
    vector<int> ms1, ms2;
    int sz = (int)v1.size();
    for (int i = 0; i <= sz - 1; i++) ok[v1[i]] = true;
    bool chk = true;
    cout << "YES\n";
    for (int i = 0; i <= n - 1; i++) {
      if (ok[a[i]] && chk) {
        cout << "A";
        chk = false;
      } else if (ok[a[i]] && !chk) {
        cout << "B";
        chk = true;
      } else
        cout << "A";
    }
    cout << "\n";
  } else {
    if (semigood) {
      cout << "YES\n";
      int sz = (int)v1.size();
      for (int i = 0; i <= sz - 1; i++) ok[v1[i]] = true;
      bool chk = true, done = true;
      for (int i = 0; i <= n - 1; i++) {
        if (ok[a[i]] && chk) {
          cout << "A";
          chk = false;
        } else if (ok[a[i]] && !chk) {
          cout << "B";
          chk = true;
        } else if (a[i] == v2[0] && done) {
          cout << "B";
          done = false;
        } else
          cout << "A";
      }
    } else
      cout << "NO\n";
  }
  return 0;
}