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 N = 1e7 + 10;
int a[N];
long long num[N];
bool isprime[N];
int prime[N], prime_cnt = 0;
int min_factor[N];
void sieve() {
  memset(isprime, 1, sizeof(isprime));
  for (int i = 2; i < N; ++i)
    if (isprime[i]) {
      min_factor[i] = i;
      prime[++prime_cnt] = i;
      if ((long long)i * i >= N) continue;
      for (int j = i * i; j < N; j += i)
        if (isprime[j]) isprime[j] = 0, min_factor[j] = i;
    }
}
inline long long cal(long long n, long long p) {
  long long ret = 0;
  for (; p <= n; n /= p) ret += n / p;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  sieve();
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int cnt = 0;
  int p = n;
  for (int i = a[n]; i >= 2; --i) {
    while (i == a[p]) cnt++, p--;
    num[i] = (long long)cnt;
  }
  for (int i = a[n]; i >= 2; --i)
    if (!isprime[i]) {
      num[min_factor[i]] += num[i];
      num[i / min_factor[i]] += num[i];
    }
  long long ans = 1;
  for (int i = 1; i <= prime_cnt && prime[i] <= a[n]; ++i) {
    long long L = 1, R = 1LL << 60;
    while (L < R) {
      long long m = (L + R) >> 1;
      if (cal(m, (long long)prime[i]) >= num[prime[i]])
        R = m;
      else
        L = m + 1;
    }
    ans = max(ans, L);
  }
  cout << ans;
}