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>
const int N = 1000000 + 5;
const int _M = 10000000 + 5;
const int _p = 670000;
int vis[_M];
int prime[_p], dep;
long long cnt[_M], INF;
void prepare() {
  vis[1] = 1;
  for (int i = 2; i < _M; ++i)
    if (!vis[i]) {
      prime[dep++] = i;
      vis[i] = i;
      if ((long long)i * i >= _M) continue;
      for (int j = i * i; j < _M; j += i) vis[j] = i;
    }
}
bool num(long long x, int y, long long z) {
  long long re = 0;
  while (x > 1) {
    x /= y;
    re += x;
    if (re >= z) return true;
  }
  return re >= z;
}
long long get(int x, long long y) {
  long long l = 0, r = INF, mid;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (num(mid, x, y))
      r = mid;
    else
      l = mid;
  }
  return r;
}
int main() {
  prepare();
  int n, a;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a);
    INF += a;
    ++cnt[a];
  }
  int b;
  for (int i = _M - 1; i >= 1; --i) cnt[i - 1] += cnt[i];
  for (int i = _M - 1; i > 1; --i)
    if (cnt[i] > 0 && vis[i] != i) {
      b = vis[i];
      cnt[b] += cnt[i];
      cnt[i / b] += cnt[i];
      cnt[i] = 0;
    }
  long long ans = 1;
  for (int i = 0; i < dep; ++i)
    if (cnt[prime[i]] > 0) ans = std::max(ans, get(prime[i], cnt[prime[i]]));
  std::cout << ans << std::endl;
  return 0;
}