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;
long long cnt[11111111];
const int maxn = 12000000;
int prime[maxn / 10], pl = 0;
bool is[maxn];
void makelist() {
  is[0] = is[1] = 1;
  int i, j;
  for (i = 2; i < maxn; ++i) {
    if (!is[i]) {
      prime[pl++] = i;
      if (1LL * i * i > 1LL * maxn) continue;
      for (j = i * i; j < maxn; j += i) is[j] = 1;
    }
  }
}
int n;
int main() {
  int i, j, k;
  scanf("%d", &n);
  int mx = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &k);
    cnt[1]++;
    cnt[k + 1]--;
    mx = max(mx, k);
  }
  for (i = 1; i <= mx; ++i) {
    cnt[i] += cnt[i - 1];
  }
  long long low = 1, high = 1LL << 61, ans, mid;
  makelist();
  for (i = 0; i < pl && prime[i] <= mx; ++i) {
    int p = prime[i];
    long long tt = 0;
    for (j = p; j <= mx; j += p) {
      int q = j, cc = 0;
      while (q % p == 0) {
        q /= p;
        cc++;
      }
      tt += (long long)cc * cnt[j];
    }
    cnt[p] = tt;
  }
  while (low <= high) {
    mid = (low + high) / 2LL;
    bool f = 1;
    for (i = 0; i < pl && prime[i] <= mx; ++i) {
      int p = prime[i];
      long long cc = 0;
      long long t = mid;
      while (t) {
        cc += t / p;
        t /= p;
      }
      if (cc < cnt[p]) {
        f = 0;
        break;
      }
    }
    if (!f) {
      low = mid + 1;
    } else {
      high = mid - 1;
      ans = mid;
    }
  }
  cout << ans << endl;
  return 0;
}