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;
const int MAX = 1e4 + 10;
const long long MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-8;
const long double PI = acos(-1);
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, x;
  cin >> n;
  vector<bool> ans(n, false);
  map<int, vector<int> > m;
  for (int i = 0; i < n; i++) {
    cin >> x;
    m[x].push_back(i);
  }
  int l = 0, p = 0;
  for (auto v : m) {
    int len = v.second.size();
    if (len == 1)
      l++;
    else if (len > 2)
      p++;
  }
  bool A = true;
  if (l % 2 == 0) {
    for (auto v : m) {
      int len = v.second.size();
      if (len == 1) {
        ans[v.second[0]] = A;
        A = !A;
      }
    }
  } else {
    bool foi = false;
    if (p == 0) {
      cout << "NO\n";
      return 0;
    }
    for (auto v : m) {
      int len = v.second.size();
      if (len == 1) {
        ans[v.second[0]] = A;
        A = !A;
      } else if (len > 2) {
        for (auto vv : v.second) {
          if (!foi) {
            ans[vv] = 0;
            foi = true;
          } else
            ans[vv] = 1;
        }
      }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << (ans[i] ? 'A' : 'B');
  cout << '\n';
  return 0;
}