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 int MOD = 1000000007;
const int INF = 100000000;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
long long powers[((int)1e7) + 10];
bool prime[((int)1e7) + 1];
void init_prime() {
  memset(prime, 1, sizeof(prime));
  prime[0] = prime[1] = false;
  for (int i = 2; i < sizeof(prime); i++)
    if (prime[i])
      for (int j = i * i; j < sizeof(prime); j += i) prime[j] = false;
}
long long get_val(int cur, long long cnt) {
  long long l = -1, r = 1LL << 60;
  while (r - l != 1) {
    long long md = (l + r) / 2;
    long long p = md;
    long long x = 0;
    while (p) {
      p /= cur;
      x += p;
    }
    if (x >= cnt)
      r = md;
    else
      l = md;
  }
  return r;
}
int main() {
  int k;
  cin >> k;
  for (int i = 0; i < (k); i++) {
    int x;
    cin >> x;
    powers[x]++;
  }
  for (int i = ((int)1e7) + 1; i >= 1; i--) powers[i] += powers[i + 1];
  long long ans = 1;
  init_prime();
  for (int i = 0; i < (((int)1e7) + 1); i++) {
    if (prime[i]) {
      long long cnt = 0;
      long long start = i;
      while (start <= ((int)1e7)) {
        for (int j = (int)start; j <= ((int)1e7); j += (int)start) {
          cnt += powers[j];
        }
        start *= i;
      }
      long long tmp = get_val(i, cnt);
      ans = max(ans, tmp);
    }
  }
  cout << ans << endl;
  return 0;
}