#include <bits/stdc++.h>
using namespace std;
auto clk = clock();
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
template <class T>
void splitStr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
inline int two(int n) { return 1 << n; }
inline int isOnBit(int n, int b) { return (n >> b) & 1; }
inline void onBit(int &n, int b) { n |= two(b); }
inline void offBit(int &n, int b) { n &= ~two(b); }
inline int lastBit(int n) { return n & (-n); }
inline int cntBit(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
const int dx4[] = {-1, 0, 1, 0};
const int dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modexp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mat1[100][100];
long long mat2[100][100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long ar[n];
  long long f[100] = {0};
  for (long long i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
    f[ar[i]]++;
  }
  for (long long i = 0, _n = (100); i < _n; i++) {
    if (f[i] != 0) {
      if (f[i] > n / 2) {
        cout << "Bob";
      } else {
        cout << "Alice";
      }
      break;
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.

Alice and Bob will play a game alternating turns with Alice going first.

On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).

Given the starting configuration, determine who will win the game.

Input

The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.

Output

Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).

Examples

Input


2
8 8


Output


Bob


Input


4
3 1 4 1


Output


Alice

Note

In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.

In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
auto clk = clock();
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
template <class T>
void splitStr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
inline int two(int n) { return 1 << n; }
inline int isOnBit(int n, int b) { return (n >> b) & 1; }
inline void onBit(int &n, int b) { n |= two(b); }
inline void offBit(int &n, int b) { n &= ~two(b); }
inline int lastBit(int n) { return n & (-n); }
inline int cntBit(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
const int dx4[] = {-1, 0, 1, 0};
const int dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modexp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mat1[100][100];
long long mat2[100][100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long ar[n];
  long long f[100] = {0};
  for (long long i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
    f[ar[i]]++;
  }
  for (long long i = 0, _n = (100); i < _n; i++) {
    if (f[i] != 0) {
      if (f[i] > n / 2) {
        cout << "Bob";
      } else {
        cout << "Alice";
      }
      break;
    }
  }
  return 0;
}

```