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 INF = 1000 * 1000 * 1000;
const int MAXV = 10000001;
bool mark[MAXV];
int lp[MAXV];
int primes[MAXV >> 1];
long long cnt[MAXV];
int psz;
void prep() {
  for (int i = 0; i < (MAXV); ++i) {
    mark[i] = true;
  }
  mark[0] = mark[1] = false;
  for (int i = 2; i < MAXV; ++i)
    if (mark[i]) {
      primes[psz++] = i;
      lp[i] = i;
      for (long long j = 2 * i; j < MAXV; j += i) {
        mark[j] = false;
        lp[j] = max(lp[j], i);
      }
    }
}
int main() {
  int N;
  scanf("%d", &N);
  long long lo = 1, hi = 0;
  for (int i = 0; i < (N); ++i) {
    int x;
    scanf("%d", &x);
    ++cnt[x];
    hi += x;
  }
  prep();
  for (int i = MAXV - 1; i > 1; --i) {
    cnt[i] += cnt[i + 1];
  }
  for (int i = MAXV - 1; i > 1; --i)
    if (lp[i] != i) {
      cnt[lp[i]] += cnt[i];
      cnt[i / lp[i]] += cnt[i];
    }
  while (lo < hi) {
    long long mid = (lo + hi) >> 1;
    bool check = true;
    for (int i = 0; i < (psz); ++i) {
      long long q = 0;
      long long tmp = mid;
      while (tmp > 0) {
        tmp /= primes[i];
        q += tmp;
      }
      if (q < cnt[primes[i]]) {
        check = false;
        break;
      }
    }
    if (!check) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  printf("%I64d", lo);
  return 0;
}