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;
long long freq[10000010ll], sieve[10000010ll];
long long m = 0;
vector<pair<long long, long long>> primes;
long long powf(long long a, long long p) {
  long long ans = 0;
  long long temp = p;
  while (temp <= a) {
    ans += a / temp;
    temp *= p;
  }
  return ans;
}
bool poss(long long n) {
  long long a, k, j;
  for (long long i = 0; i < primes.size(); i++) {
    a = powf(n, primes[i].first);
    if (a < primes[i].second) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n, sum = 0;
  cin >> n;
  for (long long i = 1; i < 10000010ll; i++) {
    freq[i] = 0;
    sieve[i] = 0;
  }
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
    freq[arr[i] + 1] -= 1;
  }
  long long mx = *max_element(arr.begin(), arr.end());
  long long ct = 0;
  freq[1] = n;
  for (long long i = 2; i <= mx + 1; i++) freq[i] += freq[i - 1];
  for (long long i = 2; i < 10000010ll; i++) {
    if (sieve[i]) continue;
    long long k = 0;
    for (long long x = i; x < 10000010ll; x *= i) {
      for (long long y = x; y < 10000010ll; y += x) {
        k += freq[y];
        if (x == i) sieve[y] = 1;
      }
    }
    m = i;
    if (k) primes.push_back({i, k});
  }
  long long l = mx, h = sum, mid = (l + h) / 2;
  while (l < h) {
    if (poss(mid))
      h = mid;
    else
      l = mid + 1;
    mid = (l + h) / 2;
  }
  cout << l << endl;
  return 0;
}