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 MAXV = 10000010;
int N;
int minP[MAXV];
int p[MAXV], pn;
long long cnt[MAXV];
long long sumP[MAXV];
void init() {
  for (int i = 2; i < MAXV; ++i) {
    if (!minP[i]) {
      minP[i] = i;
      p[pn++] = i;
    }
    for (int j = 0; j < pn && p[j] * i < MAXV; ++j) {
      minP[p[j] * i] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
int main() {
  init();
  cin >> N;
  while (N--) {
    int x;
    scanf("%d", &x);
    ++cnt[x];
  }
  for (int i = MAXV - 2; i > 1; --i) {
    cnt[i] += cnt[i + 1];
  }
  for (int i = MAXV - 2; i > 1; --i) {
    if (cnt[i]) {
      sumP[minP[i]] += cnt[i];
      cnt[i / minP[i]] += cnt[i];
    }
  }
  long long ans = 1;
  for (int i = 2; i < MAXV; ++i)
    if (sumP[i]) {
      long long low = ans - 1, high = 1LL << 50;
      while (low + 1 < high) {
        long long mid = (low + high) >> 1;
        long long t = mid, ct = 0;
        while (t) {
          ct += t / i;
          t /= i;
        }
        if (ct >= sumP[i])
          high = mid;
        else
          low = mid;
      }
      ans = high;
    }
  cout << ans << endl;
  return 0;
}