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;
int A[10000000 + 1], p[10000000 + 1], pc = 0;
void sieve() {
  for (int i = 2; i <= 10000000; i++) {
    if (!A[i]) p[A[i] = ++pc] = i;
    for (int j = 1; j <= A[i] && i * p[j] <= 10000000; j++) A[i * p[j]] = j;
  }
}
long long e[10000000 + 1];
int f[10000000 + 1];
int main() {
  sieve();
  int k;
  scanf("%d", &k);
  long long sum = 0;
  int maxA = 0;
  for (int i = 0; i < k; i++) {
    int ai;
    scanf("%d", &ai);
    sum += ai;
    maxA = max(maxA, ai);
    f[ai]++;
  }
  for (int i = 10000000; i > 0; i--) f[i - 1] += f[i];
  for (int i = 1; i <= pc; i++) {
    int pp = p[i];
    for (int j = pp; j <= 10000000; j += pp) {
      long long tmp = j, cnt = 0;
      while (tmp % pp == 0) {
        tmp /= pp;
        cnt++;
      }
      e[i] += 1LL * f[j] * cnt;
    }
  }
  long long lo = maxA - 1, hi = sum;
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    bool ok = 1;
    for (int i = 1; i <= pc; i++) {
      int pp = p[i];
      long long tmp = mid, cnt = 0;
      while (tmp) {
        tmp /= pp;
        cnt += tmp;
      }
      if (cnt < e[i]) {
        ok = 0;
        break;
      }
    }
    if (ok)
      hi = mid;
    else
      lo = mid;
  }
  cout << hi << endl;
  return 0;
}