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;
inline bool check(long long m, vector<int> &primes, vector<long long> &cnt) {
  for (int i = 0; i < (int)primes.size(); i++) {
    long long now = m, s = 0;
    while (now) {
      now /= primes[i];
      s += now;
    }
    if (s < cnt[i]) return 0;
  }
  return 1;
}
int main() {
  int n;
  while (scanf("%d", &n) >= 1) {
    vector<long long> cnt(1e7 + 2);
    vector<int> a(n);
    long long s = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      ++cnt[a[i]];
      s += a[i];
    }
    int mx = *max_element((a).begin(), (a).end());
    vector<int> getp(mx + 1, 0);
    vector<int> primes;
    for (int i = 2; i <= mx; i++) {
      if (!getp[i]) getp[i] = i, primes.push_back(i);
      for (int j = 0; j < (int)primes.size() && primes[j] <= getp[i] &&
                      (long long)i * primes[j] <= mx;
           j++)
        getp[i * primes[j]] = primes[j];
    }
    for (int i = mx - 1; i >= 2; i--) cnt[i] += cnt[i + 1];
    for (int i = mx; i >= 2; i--) {
      if (getp[i] != i) cnt[getp[i]] += cnt[i];
      cnt[i / getp[i]] += cnt[i];
    }
    for (int i = 0; i < (int)primes.size(); i++) cnt[i] = cnt[primes[i]];
    long long l = 0, r = s + 1;
    while (l < r - 1) {
      long long m = (l + r) >> 1;
      if (!check(m, primes, cnt))
        l = m;
      else
        r = m;
    }
    printf("%I64d\n", l + 1);
  }
  return 0;
}