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 asum, ppw[10000001];
bool comp[10000001];
int k, amax, a[1000000], cnt[10000001];
inline long long factpow(long long n, long long p) {
  long long ret = 0;
  while (n > 0) {
    n /= p;
    ret += n;
  }
  return ret;
}
int main() {
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
    amax = max(amax, a[i]);
    asum += a[i];
  }
  sort(a, a + k);
  int pt = 0;
  for (int i = 0; i < k; i++) {
    while (pt < a[i]) {
      pt++;
      cnt[pt] = k - i;
    }
  }
  for (int i = 2; i <= amax; i++) {
    if (comp[i]) continue;
    int pw = 1;
    while (amax / i >= pw) {
      pw *= i;
      for (int j = pw; j <= amax; j += pw) {
        comp[j] = true;
        ppw[i] += cnt[j];
      }
    }
    comp[i] = false;
  }
  long long ans = 1;
  for (int i = 2; i <= amax; i++) {
    if (ppw[i] == 0) continue;
    long long lo = 0, hi = asum;
    while (hi - lo > 1) {
      long long m = (lo + hi) / 2;
      if (factpow(m, i) >= ppw[i])
        hi = m;
      else
        lo = m;
    }
    ans = max(ans, hi);
  }
  printf("%lld\n", ans);
}