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 present[10000000 + 5];
signed long long int freq[10000000 + 5];
int prime[10000000 + 5];
void seive() {
  int N = 10000000;
  signed long long int p, temp;
  for (int i = 2; i <= N; ++i) {
    freq[i] = 0;
    if (prime[i] == 0) {
      for (int j = i * 2; j <= N; j += i) {
        prime[j] = 1;
      }
      for (signed long long int p = i; p <= N; p *= i) {
        for (signed long long int j = p; j <= N; j += p) {
          freq[i] += (present[j] - (j + p <= N ? present[j + p] : 0)) * (j / p);
        }
      }
    }
  }
}
vector<int> primes;
vector<signed long long int> powers;
bool check(signed long long int x) {
  signed long long int f = 0;
  for (int i = 0; i < primes.size(); ++i) {
    f = 0;
    for (signed long long int p = primes[i]; p <= x; p *= primes[i]) {
      f += x / p;
    }
    if (f < powers[i]) return 0;
  }
  return 1;
}
signed long long int bin(signed long long int i, signed long long int j) {
  if (i == j) return i;
  signed long long int mid = (i + j) / 2;
  if (check(mid))
    return bin(i, mid);
  else
    return bin(mid + 1, j);
}
int main() {
  int N, x, mx = 0;
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &x);
    present[x]++;
    mx = max(mx, x);
  }
  for (int i = mx; i >= 1; i--) {
    present[i] += present[i + 1];
  }
  seive();
  for (int i = 1; i <= mx; ++i) {
    if (freq[i]) {
      primes.push_back(i);
      powers.push_back(freq[i]);
    }
  }
  cout << bin(mx, 100000000000000LL);
}