Problem: In a far away galaxy there is war again. The treacherous Republic made k precision strikes of power ai on the Empire possessions. To cope with the republican threat, the Supreme Council decided to deal a decisive blow to the enemy forces. 

To successfully complete the conflict, the confrontation balance after the blow should be a positive integer. The balance of confrontation is a number that looks like <image>, where p = n! (n is the power of the Imperial strike), <image>. After many years of war the Empire's resources are low. So to reduce the costs, n should be a minimum positive integer that is approved by the commanders.

Help the Empire, find the minimum positive integer n, where the described fraction is a positive integer.

Input

The first line contains integer k (1 ≤ k ≤ 106). The second line contains k integers a1, a2, ..., ak (1 ≤ ai ≤ 107).

Output

Print the minimum positive integer n, needed for the Empire to win.

Please, do not use the %lld to read or write 64-but integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2
1000 1000


Output

2000

Input

1
2


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long nmax = 1e7 + 5;
long long cnt[nmax];
bool prime[nmax];
vector<long long> primes;
vector<long long> cost;
long long f(long long n, long long p) {
  if (n < p) return 0;
  return (n / p) + f(n / p, p);
}
bool solve(long long n) {
  for (long long i = 0; i < primes.size(); ++i)
    if (f(n, primes[i]) < cost[i]) {
      return false;
    }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long k;
  cin >> k;
  for (long long i = 0; i < k; ++i) {
    long long a;
    cin >> a;
    cnt[a]++;
  }
  for (long long i = nmax - 1; i >= 0; --i) cnt[i] += cnt[i + 1];
  for (long long i = 2; i < nmax; ++i) prime[i] = true;
  for (long long i = 0; i < nmax; ++i)
    if (prime[i]) {
      primes.push_back(i);
      for (long long j = 2 * i; j < nmax; j += i) prime[j] = false;
    }
  cost.resize(primes.size());
  for (long long i = 0; i < primes.size(); ++i) {
    long long p = primes[i];
    while (true) {
      for (long long j = p; j < nmax; j += p) cost[i] += cnt[j];
      p *= primes[i];
      if (p >= nmax) break;
    }
  }
  long long lo = 1, hi = 1e15, mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (solve(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  cout << lo << '\n';
}