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 long long inf = 2147383647;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
int freq[200];
int ans[200];
int main() {
  int n;
  cin >> n;
  int x;
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    cin >> x;
    v.push_back({x, i});
    freq[x]++;
  }
  int cntA = 0, cntB = 0;
  bool flag = true;
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    if (freq[v[i].first] == 1) {
      if (flag) {
        ans[v[i].second] = 1;
        cntA++;
        flag = false;
      } else {
        ans[v[i].second] = -1;
        cntB++;
        flag = true;
      }
    }
  }
  for (int i = 0; i < v.size(); i++) {
    if (freq[v[i].first] == 2) {
      ans[v[i].second] = 1;
      ans[v[i + 1].second] = -1;
      i++;
    }
  }
  for (int i = 0; i < v.size(); i++) {
    if (freq[v[i].first] > 2) {
      int temp = v[i].first;
      if (cntA < cntB) {
        ans[v[i].second] = 1;
        i = i + 1;
        while (v[i].first == temp) {
          ans[v[i].second] = -1;
          i++;
        }
        i--;
        cntA++;
      } else if (cntB < cntA) {
        ans[v[i].second] = -1;
        i = i + 1;
        while (v[i].first == temp) {
          ans[v[i].second] = 1;
          i++;
        }
        i--;
        cntB++;
      } else {
        while (v[i].first == temp) {
          ans[v[i].second] = 1;
          i++;
        }
        i--;
      }
    }
  }
  if (cntA == cntB) {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      if (ans[i] == 1)
        cout << "A";
      else
        cout << "B";
    }
  } else
    cout << "NO" << endl;
}