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;
using ll = long long;
const int N = 1000006;
const int M = 10000007;
int k, a[N];
int lp[M];
ll cnt[M];
int p[M], sz;
bool f(ll x) {
  for (int i = 0; i < sz; i++) {
    ll tmp = x;
    ll fac = 0;
    while (tmp) {
      tmp /= p[i];
      fac += tmp;
    }
    if (fac < cnt[p[i]]) return false;
  }
  return true;
}
int main() {
  for (int i = 2; i * i < M; i++)
    if (!lp[i]) {
      for (int j = i * i; j < M; j += i)
        if (!lp[j]) {
          lp[j] = i;
        }
    }
  for (int i = 2; i < M; i++) {
    if (!lp[i]) lp[i] = i;
    if (lp[i] == i) p[sz++] = i;
  }
  scanf("%d", &k);
  for (int i = 0; i < k; i++) scanf("%d", a + i);
  sort(a, a + k);
  for (int i = 0; i < k; i++) cnt[a[i]]++;
  for (int i = M - 2; i >= 0; i--) cnt[i] += cnt[i + 1];
  for (int i = M - 1; i >= 2; i--)
    if (lp[i] != i and cnt[i]) {
      int x = lp[i], b = i / x;
      cnt[x] += cnt[i];
      cnt[b] += cnt[i];
      cnt[i] = 0;
    }
  while (sz > 0 and cnt[p[sz - 1]] == 0) sz--;
  ll lo = 1, hi = 1e16, ans = -1;
  while (lo <= hi) {
    ll mid = (lo + hi) / 2;
    if (f(mid))
      ans = mid, hi = mid - 1;
    else
      lo = mid + 1;
  }
  printf("%lld\n", ans);
}