#include <bits/stdc++.h>
using namespace std;
const long long mx = 200001;
const long long mod = 1e9 + 7;
const long double PI = 3.141592653589793238460;
void __print(long long x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<int> rng(0, 1);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return ((long long)ceil(log2(n)) == (long long)floor(log2(n)));
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % mod) * powm(b, mod - 2) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i * i <= n; i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long fact[mx];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], mod - 2));
  z = mul(z, powm(fact[r], mod - 2));
  return z;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long x = 0;
  while (x < n - 1 && a[x + 1] > a[x]) x++;
  long long y = n - 1;
  while (y > 0 && a[y - 1] > a[y]) y--;
  x++;
  y = n - y;
  if (x % 2 == 0 && y % 2 == 0)
    cout << "Bob\n";
  else
    cout << "Alice\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(20);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Alice and Bob are playing a game. They are given an array A of length N. The array consists of integers. They are building a sequence together. In the beginning, the sequence is empty. In one turn a player can remove a number from the left or right side of the array and append it to the sequence. The rule is that the sequence they are building must be strictly increasing. The winner is the player that makes the last move. Alice is playing first. Given the starting array, under the assumption that they both play optimally, who wins the game?

Input

The first line contains one integer N (1 ≤ N ≤ 2*10^5) - the length of the array A.

The second line contains N integers A_1, A_2,...,A_N (0 ≤ A_i ≤ 10^9)

Output

The first and only line of output consists of one string, the name of the winner. If Alice won, print "Alice", otherwise, print "Bob".

Examples

Input


1
5


Output


Alice


Input


3
5 4 5


Output


Alice


Input


6
5 8 2 1 10 9


Output


Bob

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mx = 200001;
const long long mod = 1e9 + 7;
const long double PI = 3.141592653589793238460;
void __print(long long x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<int> rng(0, 1);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return ((long long)ceil(log2(n)) == (long long)floor(log2(n)));
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % mod) * powm(b, mod - 2) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i * i <= n; i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long fact[mx];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], mod - 2));
  z = mul(z, powm(fact[r], mod - 2));
  return z;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long x = 0;
  while (x < n - 1 && a[x + 1] > a[x]) x++;
  long long y = n - 1;
  while (y > 0 && a[y - 1] > a[y]) y--;
  x++;
  y = n - y;
  if (x % 2 == 0 && y % 2 == 0)
    cout << "Bob\n";
  else
    cout << "Alice\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(20);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```