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 PI = acos((long double)-1.0);
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  int idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ",";
  dbs(str.substr(idx + 1), s...);
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream& operator<<(ostream& os, const map<S, T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
vector<long long> a, b;
const long long U = 1e18;
void recurse(int index1, long long currNum, vector<int>& temp,
             vector<long long>& a) {
  if (index1 == int(temp.size())) {
    a.push_back(currNum);
    return;
  }
  recurse(index1 + 1, currNum, temp, a);
  while (currNum <= U / (long long)temp[index1]) {
    currNum *= (long long)temp[index1];
    recurse(index1 + 1, currNum, temp, a);
  }
}
void precompute(vector<int>& temp) {
  vector<int> temp1, temp2;
  for (int i = (int)(0); i <= (int)(int(temp.size()) - 1); i++) {
    if (i % 2 == 0)
      temp1.push_back(temp[i]);
    else
      temp2.push_back(temp[i]);
  }
  recurse(0, 1LL, temp1, a);
  recurse(0, 1LL, temp2, b);
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  reverse(a.begin(), a.end());
}
bool possible(long long val1, long long lowerThan) {
  int j = 0;
  long long answer1 = 0;
  for (int i = (int)(0); i <= (int)(int(a.size()) - 1); i++) {
    while (j < int(b.size()) && b[j] <= val1 / a[i]) j++;
    answer1 += (long long)j;
  }
  if (answer1 < lowerThan)
    return false;
  else
    return true;
}
long long solve(long long k) {
  long long low1 = 1;
  long long high1 = 1e18;
  while (low1 < high1) {
    if (low1 + 1 == high1) {
      if (possible(low1, k))
        high1 = low1;
      else
        low1 = high1;
      break;
    }
    long long mid = (low1 + high1) / 2LL;
    if (possible(mid, k))
      high1 = mid;
    else
      low1 = mid + 1;
  }
  return low1;
}
int main() {
  int n;
  cin >> n;
  vector<int> v;
  for (int i = (int)(0); i <= (int)(n - 1); i++) {
    int temp;
    cin >> temp;
    v.push_back(temp);
  }
  precompute(v);
  long long k;
  cin >> k;
  cout << solve(k) << "\n";
  return 0;
}