Problem: Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set. 

Input

The first line contains a single integer n (1 ≤ n ≤ 16).

The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.

The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.

Output

Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.

Examples

Input

3
2 3 5
7


Output

8


Input

5
3 7 11 13 31
17


Output

93

Note

The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:

(1, 2, 3, 4, 5, 6, 8, ...)

The seventh number in this list (1-indexed) is eight.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-10;
const long long MOD = 1e9 + 7;
const long double PI = 3.1415926535897932384626433832795028841;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
ostream& operator<<(std::ostream& stream, const long long* arr) {
  int i = 0;
  cout << "{";
  while (arr[i] != INT_MAX) {
    cout << arr[i];
    if (arr[i + 1] != INT_MAX) cout << ",";
    i++;
  }
  cout << "}";
  return stream;
}
ostream& operator<<(std::ostream& stream, const vector<int> vec) {
  cout << "{";
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i];
    if (i != vec.size() - 1) cout << ",";
  }
  cout << "}";
  return stream;
}
ostream& operator<<(std::ostream& stream,
                    const pair<long long, long long>& vec) {
  cout << "(" << vec.first << ',' << vec.second << ")";
  return stream;
}
template <class T>
struct segtree_t {
  vector<T> t, lazy, flag_equal;
  int st, sz;
  T (*addition_fun)(T, T);
  T invalid_value;
  segtree_t() {}
  segtree_t(int start, int finish, T* arr, T (*add_fun)(T, T), T invalid_val) {
    int size = (finish + 1) * 2;
    int tree_size = 2;
    while (tree_size < size) tree_size *= 2;
    t = vector<T>(tree_size);
    lazy = vector<T>(tree_size);
    flag_equal = vector<T>(tree_size);
    st = start;
    sz = finish;
    addition_fun = add_fun;
    invalid_value = invalid_val;
    build(1, start, finish, arr);
  }
  void build(int i, int l, int r, T* arr) {
    if (l > r) return;
    lazy[i] = 0;
    flag_equal[i] = 0;
    if (l == r) {
      t[i] = arr[l];
    } else {
      build(i * 2, l, (l + r) / 2, arr);
      build(i * 2 + 1, (l + r) / 2 + 1, r, arr);
      t[i] = addition_fun(t[i * 2], t[i * 2 + 1]);
    }
  }
  void inc_range(int l, int r, T value) {
    if (l > r) return;
    inc_range_rec(1, st, sz, l, r, value);
  }
  void set_values(int l, int r, T value) {
    set_values_rec(1, st, sz, l, r, value);
  }
  void inc_range_rec(int i, int l, int r, int tl, int tr, T value) {
    if (lazy[i] != 0) {
      t[i] += lazy[i];
      if (l != r) {
        lazy[i * 2] += lazy[i];
        lazy[i * 2 + 1] += lazy[i];
      }
      lazy[i] = 0;
    }
    if (r < tl or l > tr) return;
    if (l > r) return;
    if (l >= tl and r <= tr) {
      if (l != r) {
        lazy[i * 2] += value;
        lazy[i * 2 + 1] += value;
      }
      t[i] += value;
    } else {
      inc_range_rec(i * 2, l, (l + r) / 2, tl, tr, value);
      inc_range_rec(i * 2 + 1, (l + r) / 2 + 1, r, tl, tr, value);
      t[i] = addition_fun(t[i * 2], t[i * 2 + 1]);
    }
  }
  T query_rec(int i, int l, int r, int tl, int tr) {
    if (flag_equal[i] != 0) {
      if (l != r) {
        t[i * 2] = t[i];
        t[i * 2 + 1] = t[i];
        flag_equal[i * 2] = 1;
        flag_equal[i * 2 + 1] = 1;
      }
      flag_equal[i] = 0;
    }
    if (r < tl or l > tr) return invalid_value;
    if (l > r) return invalid_value;
    if (lazy[i] != 0) {
      t[i] += lazy[i];
      if (l != r) {
        lazy[i * 2] += lazy[i];
        lazy[i * 2 + 1] += lazy[i];
      }
      lazy[i] = 0;
    }
    if (l >= tl and r <= tr) {
      return t[i];
    } else {
      T m1 = query_rec(i * 2, l, (l + r) / 2, tl, tr);
      T m2 = query_rec(i * 2 + 1, (l + r) / 2 + 1, r, tl, tr);
      if (m1 == invalid_value && m2 == invalid_value) return invalid_value;
      if (m1 == invalid_value) return m2;
      if (m2 == invalid_value) return m1;
      return addition_fun(m1, m2);
    }
  }
  T query(int l, int r) { return query_rec(1, st, sz, l, r); }
  void set_values_rec(int i, int l, int r, int tl, int tr, T value) {
    if (flag_equal[i] != 0) {
      if (l != r) {
        t[i * 2] = t[i];
        t[i * 2 + 1] = t[i];
        flag_equal[i * 2] = 1;
        flag_equal[i * 2 + 1] = 1;
      }
      flag_equal[i] = 0;
    }
    if (r < tl || l > tr) return;
    if (l > r) return;
    if (l >= tl and r <= tr) {
      t[i] = value;
      flag_equal[i] = 1;
    } else {
      set_values_rec(i * 2, l, (l + r) / 2, tl, tr, value);
      set_values_rec(i * 2 + 1, (l + r) / 2 + 1, r, tl, tr, value);
      t[i] = addition_fun(t[i * 2], t[i * 2 + 1]);
    }
  }
  ~segtree_t() {}
};
template <class T>
struct fenwick_t {
  vector<T> FW;
  fenwick_t(int n) { FW = vector<T>(n + 1); }
  void inc(int i, T v) {
    while (i <= FW.size() - 1) {
      FW[i] += v;
      i += i & (-i);
    }
  }
  T get(int i) {
    T rta = 0;
    while (i > 0) {
      rta += FW[i];
      i -= i & (-i);
    }
    return rta;
  }
};
long long n;
vector<long long> primes_a, primes_b;
vector<long long> Ag, Bg;
void generate(vector<long long>& primes, vector<long long>& group) {
  if (primes.size() == 0) return;
  queue<pair<long long, long long> > Q;
  Q.push(pair<long long, long long>(1, 0));
  while (!Q.empty()) {
    pair<long long, long long> act =
        pair<long long, long long>(Q.front().first, Q.front().second);
    Q.pop();
    group.push_back(act.first);
    if (act.first < 1e18 / primes[act.second] + 1)
      Q.push(pair<long long, long long>(act.first * primes[act.second],
                                        act.second));
    for (int i = act.second + 1; i < primes.size(); i++) {
      if (act.first < 1e18 / primes[i] + 1)
        Q.push(pair<long long, long long>(act.first * primes[i], i));
    }
  }
}
long long count_numbers(long long h) {
  if (Bg.size() == 0) {
    long long A = -1;
    long long B = Ag.size();
    while (A + 1 != B) {
      long long half = (A + B) / 2;
      if (Ag[half] >= h) {
        B = half;
      } else {
        A = half;
      }
    }
    return Ag[B] == h ? B + 1 : B;
  }
  long long var_b = Bg.size() - 1;
  long long count = 0;
  for (long long a = 0; a < Ag.size(); a++) {
    while (var_b >= 0 &&
           (long double)(Ag[a]) > (long double)(h) / (long double)(Bg[var_b]))
      var_b--;
    count += var_b + 1;
  }
  return count;
}
class TaskE {
 public:
  void solve(std::istream& in, std::ostream& out) {
    ios::sync_with_stdio(0);
    in >> n;
    primes_a.clear();
    primes_b.clear();
    for (long long i = 0; i < n; i++) {
      long long v;
      in >> v;
      if (i % 2 == 0) primes_a.push_back(v);
      if (i % 2 == 1) primes_b.push_back(v);
    }
    Ag.clear();
    Bg.clear();
    generate(primes_a, Ag);
    generate(primes_b, Bg);
    sort(Ag.begin(), Ag.end());
    sort(Bg.begin(), Bg.end());
    long long k;
    in >> k;
    long long A = 0;
    long long B = 1e18 + 1;
    while (A + 1 != B) {
      long long h = (A + B) / 2;
      if (count_numbers(h) >= k) {
        B = h;
      } else {
        A = h;
      }
    }
    out << B << '\n';
  }
};
int main() {
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}