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 MAX = int(1e7) + 10;
int K, M;
long long cnt[MAX];
long long lo = 1, hi = 0;
int maxP[MAX];
vector<int> primes;
void sieve() {
  for (int i = 2; i < MAX; i++)
    if (!maxP[i]) {
      primes.push_back(i);
      for (int j = i; j < MAX; j += i) maxP[j] = i;
    }
}
long long calc(long long n, long long p) {
  long long ret = 0, pot = p;
  while (pot <= n) {
    ret += n / pot;
    pot *= p;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  sieve();
  cin >> K;
  for (int i = 0; i < K; i++) {
    int a;
    cin >> a;
    cnt[a]++;
    M = max(M, a);
    hi += a;
  }
  for (int i = M; i > 1; i--) cnt[i - 1] += cnt[i];
  for (int i = M; i > 1; i--) {
    if (maxP[i] == i) continue;
    cnt[maxP[i]] += cnt[i];
    cnt[i / maxP[i]] += cnt[i];
  }
  for (; lo < hi;) {
    long long mit = (lo + hi) >> 1;
    bool flag = true;
    for (int i = 0; i < (int)primes.size() && flag; i++)
      if (calc(mit, primes[i]) < cnt[primes[i]]) flag = false;
    if (!flag)
      lo = mit + 1;
    else
      hi = mit;
  }
  cout << lo << endl;
  return 0;
}