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 MAXN = 1e6 + 20, MAXV = 1e7 + 20;
int n;
long long a[MAXN], minf[MAXV], rc[MAXV];
vector<long long> prime;
map<long long, long long> c;
void sieve(long long ub) {
  for (long long i = 2; i <= ub; i++) {
    if (!minf[i]) {
      minf[i] = i;
      prime.push_back(i);
    }
    for (long long p : prime) {
      if (minf[i] < p || i * p > ub) break;
      minf[i * p] = p;
    }
  }
}
bool check(long long n) {
  for (const auto& [p, k] : c) {
    long long tmp = n, cur_k = 0;
    while (tmp) cur_k += (tmp /= p);
    if (k > cur_k) return false;
  }
  return true;
}
int main() {
  memset(minf, 0, sizeof minf);
  memset(rc, 0, sizeof rc);
  long long max_a = 0, sum_a = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++) {
    max_a = max(max_a, a[i]);
    sum_a += a[i];
    rc[a[i]]++;
  }
  sieve(max_a);
  for (long long i = max_a; i >= 2; i--) rc[i] += rc[i + 1];
  for (long long i = max_a; i >= 2; i--) {
    if (minf[i] != i) rc[minf[i]] += rc[i];
    rc[i / minf[i]] += rc[i];
  }
  for (long long p : prime) c[p] = rc[p];
  long long l = 1, r = sum_a;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  printf("%lld\n", l);
}