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 + 2, Maxv = 1e7 + 2;
int top, prim[Maxv];
int notPrim[Maxv];
void getPrim() {
  for (int i = 2; i < Maxv; ++i) {
    if (!notPrim[i]) prim[top++] = i;
    for (int j = 0; j < top && prim[j] * i < Maxv; ++j) {
      notPrim[prim[j] * i] = i;
      if (i % prim[j] == 0) break;
    }
  }
}
int n, a[Maxn];
long long cnt[Maxv];
int large;
long long getSum(long long v, int p) {
  long long ret = 0;
  while (v) {
    ret += v / p;
    v /= p;
  }
  return ret;
}
bool check(long long v) {
  for (int i = 0; i < top && prim[i] <= large; ++i) {
    int p = prim[i];
    if (getSum(v, p) < cnt[p]) return false;
  }
  return true;
}
int main() {
  getPrim();
  scanf("%d", &n);
  large = 0;
  long long r = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
    large = max(large, a[i]);
    r += a[i];
  }
  for (int i = large - 1; i >= 1; --i) cnt[i] += cnt[i + 1];
  for (int i = large; i >= 1; --i) {
    if (notPrim[i]) {
      cnt[notPrim[i]] += cnt[i];
      cnt[i / notPrim[i]] += cnt[i];
      cnt[i] = 0;
    }
  }
  long long l = 1;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%I64d\n", l);
  return 0;
}