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>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1.0);
const long double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 100;
const int N = 2e5 + 100;
const int M = (1 << 20) + 10;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  x = f ? -x : x;
}
template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  T y = 1;
  int len = 1;
  for (; y <= x / 10; y *= 10) ++len;
  for (; len; --len, x %= y, y /= 10) putchar(x / y + 48);
}
long long qpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
const long long Mx = 1e18;
long long n, k;
vector<long long> a[2], v[2];
void dfs(int id, long long cur, int pos, long long mx) {
  if (pos == a[id].size()) {
    v[id].push_back(cur);
    return;
  }
  long long val = 1, tmp = a[id][pos];
  while (1) {
    dfs(id, cur * val, pos + 1, mx);
    if (1.0 * cur * val * tmp > 1.0 * mx) break;
    val *= tmp;
  }
}
long long check(long long mid) {
  long long res = 0;
  int j = 0;
  for (int i = v[0].size() - 1; i >= 0; --i) {
    while (j < v[1].size() && v[1][j] <= mid / v[0][i]) ++j;
    res += j;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    a[i & 1].push_back(x);
  }
  cin >> k;
  dfs(0, 1, 0, Mx);
  dfs(1, 1, 0, Mx);
  sort(v[0].begin(), v[0].end());
  sort(v[1].begin(), v[1].end());
  long long ans = -1;
  long long l = 1, r = 1e18;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid) >= k)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans << "\n";
  return 0;
}