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>
T abs(T n) {
  return (n < 0 ? -n : n);
}
template <typename T>
T sqr(T n) {
  return (n * n);
}
int n;
vector<long long> lp, pr, a;
long long st[(int)1e7 + 10];
long long mx = 0;
void build() {
  for (int i = 2; i <= mx; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0;
         j < (int)pr.size() && pr[j] <= lp[i] && i * 1ll * pr[j] <= mx; ++j)
      lp[pr[j] * i] = pr[j];
  }
}
long long fact_pow(long long i, long long m) {
  long long res = 0;
  while (m) {
    m /= i;
    res += m;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    sum += a[i];
    mx = max(mx, a[i]);
  }
  lp.resize(mx + 1);
  build();
  a.push_back(0);
  sort(a.begin(), a.end());
  for (int i = 1; i <= n; i++) {
    for (int j = a[i - 1] + 1; j <= a[i]; j++) {
      int x = j;
      while (x > 1) {
        st[lp[x]] += (n - i + 1);
        x /= lp[x];
      }
    }
  }
  long long l = 1, r = sum, ans = r;
  while (l <= r) {
    long long m = (l + r) >> 1;
    bool flag = true;
    for (int i = 2; i <= (int)1e7; i++) {
      if (st[i] > 0 && fact_pow(i, m) < st[i]) flag = false;
    }
    if (!flag)
      l = m + 1;
    else {
      ans = min(ans, m);
      r = m - 1;
    }
  }
  cout << r + 1;
  return 0;
}