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;
template <typename T>
inline void checkMin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
inline void checkMax(T& a, T b) {
  if (a < b) a = b;
}
const int MAXN = (int)(1E7 + 7);
vector<int> p;
int lp[MAXN];
long long cp[MAXN];
int a[MAXN];
inline bool check(long long n) {
  for (auto i : p) {
    if (i > n) {
      return true;
    }
    long long c = 0;
    for (long long x = n / i; x > 0; x /= i) {
      c += x;
    }
    if (c < cp[i]) {
      return false;
    }
  }
  return true;
}
int main() {
  for (int i = 2; i < MAXN; ++i) {
    if (lp[i] == 0) {
      p.push_back(i);
      for (int j = i; j < MAXN; j += i) {
        lp[j] = i;
      }
    }
  }
  int k, x, y = 0;
  long long l, r = 0;
  cin >> k;
  for (int i = 0; i < int(k); ++i) {
    cin >> x;
    ++a[x];
    checkMax(y, x);
    r += x;
  }
  l = y;
  for (k = 0; y >= 2; --y) {
    for (x = y, k += a[y]; x > 1; x /= lp[x]) {
      cp[lp[x]] += k;
    }
  }
  while (l < r) {
    if (check((l + r) / 2)) {
      r = (l + r) / 2;
    } else {
      l = (l + r) / 2 + 1;
    }
  }
  cout << l << endl;
  return 0;
}