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 debug = 0;
void pr_v(vector<vector<int>> &v) {
  for (int i = 0; i < (int)v.size(); ++i) {
    cout << i << ": ";
    for (int j = 0; j < (int)v[i].size(); ++j) cout << v[i][j] << " ";
    cout << '\n';
  }
}
void pr_v(vector<pair<long long, long long>> &v) {
  for (int i = 0; i < (int)v.size(); ++i)
    cout << "{" << v[i].first << ", " << v[i].second << "} ";
  cout << '\n';
}
void pr_v(vector<char> &v) {
  for (int i = 0; i < (int)v.size(); ++i) cout << v[i] << " ";
  cout << '\n';
}
int main() {
  if (!debug) ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first;
    v[i].second = i;
  }
  sort(v.begin(), v.end());
  if (debug)
    for (int i = 0; i < n; ++i) cout << v[i].first << " ";
  vector<char> ans(n, '?');
  vector<int> inds;
  inds.push_back(v[0].second);
  int ans1 = 0, ans2 = 0, any = 0, kolnow = 1, now = v[0].first;
  for (int i = 1; i < n; ++i) {
    if (v[i].first != now) {
      if (kolnow == 1) {
        if (ans1 == ans2) {
          ans[v[i - 1].second] = 'A';
          ++ans1;
        } else {
          ans[v[i - 1].second] = 'B';
          ++ans2;
        }
      } else if (kolnow == 2) {
        ans[inds[0]] = 'A';
        ans[inds[1]] = 'B';
      } else if (kolnow >= 3)
        ++any;
      if (debug) cout << "///" << kolnow << '\n';
      inds.clear();
      kolnow = 1;
      now = v[i].first;
      inds.push_back(v[i].second);
      if (debug) cout << i << '\n';
    } else {
      inds.push_back(v[i].second);
      ++kolnow;
    }
  }
  if (kolnow == 1) {
    if (ans1 == ans2) {
      ans[v[n - 1].second] = 'A';
      ++ans1;
    } else {
      ans[v[n - 1].second] = 'B';
      ++ans2;
    }
  } else if (kolnow == 2) {
    ans[inds[0]] = 'A';
    ans[inds[1]] = 'B';
  } else if (kolnow >= 3)
    ++any;
  if (debug) cout << ans1 << " " << ans2 << '\n';
  if (ans1 - ans2 > any) return cout << "NO", 0;
  if (debug) pr_v(ans);
  if (ans1 > ans2)
    for (int i = 0; i < n; ++i)
      if (ans[i] == '?') {
        ans[i] = 'B';
        break;
      }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) cout << (ans[i] == '?' ? 'A' : ans[i]);
  return 0;
}