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 std::cin;
using std::cout;
int n, m, k, cnt[10000000 + 1], psum[10000000 + 1], a[1000000 + 1], mx,
    pr[10000000 + 1];
long long ans;
std::vector<int> pv;
std::vector<long long> need;
int max(int a, int b) { return (a < b) ? b : a; }
void in() {
  cin >> k;
  mx = 0;
  for (int i = 0; i < k; ++i) cin >> a[i], ++cnt[a[i]], mx = max(mx, a[i]);
}
void sprime() {
  for (int i = 2; i <= mx; ++i)
    if (pr[i] == 0) {
      pr[i] = i;
      pv.push_back(i);
      for (int j = i + i; j <= mx; j += i) pr[j] = i;
    }
}
void sfact() {
  need.resize(pv.size());
  for (int i = 0; i < pv.size(); ++i) {
    int p = pv[i];
    for (int j = p;; j *= p) {
      for (int l = j, c = 1; l <= mx; l += j, ++c)
        need[i] +=
            (long long)c *
            (((l + j - 1 > mx) ? psum[mx] : psum[l + j - 1]) - psum[l - 1]);
      if ((long long)j * p > (long long)mx) break;
    }
  }
}
bool check(long long x) {
  for (int i = 0; i < pv.size(); ++i) {
    long long sum = 0;
    int p = pv[i];
    for (long long j = p; j <= x; j *= p) {
      sum += x / j;
    }
    if (sum < need[i]) return 0;
  }
  return 1;
}
long long bsearch() {
  long long mn = 0;
  for (int i = 0; i < k; ++i) mn += a[i];
  long long f = mx, l = mn;
  while (f <= l) {
    long long mid = (f + l) >> 1;
    if (check(mid)) {
      mn = mid;
      l = mid - 1;
    } else
      f = mid + 1;
  }
  return mn;
}
void prs() {
  for (int i = 1; i <= mx; ++i) psum[i] = psum[i - 1] + cnt[i];
  sprime();
  sfact();
  ans = bsearch();
}
void out() { cout << ans; }
void run() {
  in();
  prs();
  out();
}
int main() {
  run();
  return 0;
}