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 long int maxn = 1e18;
int prime[20];
long long int k;
vector<int> a, b;
vector<long long> A, B;
void dfs1(long long int val, int ind) {
  if (ind == (int)a.size()) {
    A.push_back(val);
    return;
  }
  dfs1(val, ind + 1);
  if ((maxn / val) >= 1ll * a[ind]) dfs1(val * a[ind], ind);
}
void dfs2(long long int val, int ind) {
  if (ind == (int)b.size()) {
    B.push_back(val);
    return;
  }
  dfs2(val, ind + 1);
  if ((maxn / val) >= 1ll * b[ind]) dfs2(val * b[ind], ind);
}
long long int find_Upper1(long long int l, long long int r, long long int val) {
  if (l <= r) {
    long long int m = l + (r - l) / 2;
    if (B[m] <= val and (m == r or B[m + 1] > val)) return (m + 1);
    if (B[m] <= val) return find_Upper1(m + 1, r, val);
    return find_Upper1(l, m - 1, val);
  }
}
long long int find_Upper2(long long int l, long long int r, long long int val) {
  if (l <= r) {
    long long int m = l + (r - l) / 2;
    if (A[m] <= val and (m == r or A[m + 1] > val)) return (m + 1);
    if (A[m] <= val) return find_Upper2(m + 1, r, val);
    return find_Upper2(l, m - 1, val);
  }
}
long long int cal(long long int val) {
  long long int sum = 0;
  if ((int)B.size() > (int)A.size()) {
    for (int i = 0; i <= (int)A.size() - 1; i++) {
      if (A[i] > val) break;
      sum += find_Upper1(0, (int)B.size() - 1, val / A[i]);
    }
  } else {
    for (int i = 0; i <= (int)B.size() - 1; i++) {
      if (B[i] > val) break;
      sum += find_Upper2(0, (int)A.size() - 1, val / B[i]);
    }
  }
  return sum;
}
long long int srch(long long int start, long long int end) {
  if (end >= start) {
    long long int mid = start + (end - start) / 2;
    long long int x = cal(mid);
    if (x == k and (mid == start or cal(mid - 1) < k)) return mid;
    if (x < k) return srch(mid + 1, end);
    return srch(start, mid - 1);
  }
  return 0;
}
void solve() {
  int n, p, z, y, x;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &prime[i]);
  scanf("%lld", &k);
  for (int i = 1; i <= n; i++) {
    if (i & 1)
      a.push_back(prime[i]);
    else
      b.push_back(prime[i]);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  sort((A).begin(), (A).end());
  sort((B).begin(), (B).end());
  long long int low = 1, high = 1e18;
  printf("%lld\n", srch(low, high));
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}