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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int *a = new int[n];
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  int nice = 0;
  for (auto it : mp) {
    if (it.second == 1) nice++;
  }
  bool v[105];
  for (int i = 0; i < 105; i++) v[i] = false;
  bool vis[n];
  for (int i = 0; i < n; i++) vis[i] = false;
  char b[n];
  int x, w = 0;
  if (nice % 2 == 0) {
    for (int i = 0; i < n && w < nice / 2; i++) {
      if (mp[a[i]] == 1 && v[a[i]] == false) {
        w++;
        b[i] = 'A';
        v[a[i]] = true;
        vis[i] = true;
      }
    }
    w = 0;
    for (int i = 0; i < n && w < nice / 2; i++) {
      if (mp[a[i]] == 1 && v[a[i]] == false) {
        w++;
        b[i] = 'B';
        v[a[i]] = true;
        vis[i] = true;
      }
    }
  } else {
    bool ok = false;
    for (auto it : mp) {
      if (it.second > 2) {
        ok = true;
        x = it.first;
        break;
      }
    }
    if (!ok) {
      cout << "NO";
      return 0;
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == x && vis[i] == false) {
        vis[i] = true;
        b[i] = 'A';
        break;
      }
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == x && vis[i] == false) {
        vis[i] = true;
        b[i] = 'B';
      }
    }
    w = 0;
    for (int i = 0; i < n && w < nice / 2; i++) {
      if (mp[a[i]] == 1 && v[a[i]] == false) {
        w++;
        v[a[i]] = true;
        vis[i] = true;
        b[i] = 'A';
      }
    }
    w = 0;
    for (int i = 0; i < n && w < nice - nice / 2; i++) {
      if (mp[a[i]] == 1 && v[a[i]] == false) {
        w++;
        v[a[i]] = true;
        vis[i] = true;
        b[i] = 'B';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (vis[i] == false) b[i] = 'A';
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << b[i];
  return 0;
}