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;
int num[1 << 24];
long long fre[1 << 24];
void precompute(void) {
  int j;
  num[0] = num[1] = 1;
  for (int i = 2; i * i <= 11000000; i += 1) {
    if (num[i] == 0) {
      for (j = i; j <= 11000000; j += i) num[j] = i;
    }
  }
  for (int i = 2; i <= 11000000; i++) {
    if (num[i] == 0 || num[i] == i) {
      num[i] = i;
    }
  }
}
long long avail(long long x, long long a) {
  long long ret = 0;
  while (x) {
    x /= a;
    ret += x;
  }
  return ret;
}
int main() {
  long long k, x;
  long long i;
  precompute();
  cin >> k;
  for (i = 1; i <= k; i++) {
    cin >> x;
    fre[x]++;
  }
  for (i = 11000000 - 1; i >= 1; i--) fre[i] += fre[i + 1];
  for (i = 11000000 - 1; i >= 2; i--) {
    if (fre[i] == 0) continue;
    if (num[i] != i) {
      fre[i / num[i]] += fre[i];
      fre[num[i]] += fre[i];
      fre[i] = 0;
    }
  }
  long long ans, bignum, low, high, mid;
  ans = 1;
  bignum = 10000000000000LL;
  for (i = 2; i <= 11000000; i++) {
    if (fre[i] > avail(ans, i)) {
      low = ans + 1;
      high = bignum;
      while (low <= high) {
        mid = (low + high) / 2;
        if (avail(mid, i) >= fre[i])
          ans = mid, high = mid - 1;
        else
          low = mid + 1;
      }
      if (avail(ans, i) < fre[i]) {
        while (1)
          ;
        exit(0);
      }
    }
  }
  cout << ans << endl;
  return 0;
}