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;
const int MAXM = 1e7 + 5;
bool pri[MAXM];
int minP[MAXM], k, prim[MAXM], tot;
long long sum, num[MAXM];
long long val[MAXM];
void getPrim() {
  memset(pri, 1, sizeof(pri));
  for (int i = 2; i < MAXM; i++) {
    if (pri[i]) {
      minP[i] = i, prim[tot++] = i;
      for (int j = 2 * i; j < MAXM; j += i) {
        if (pri[j]) minP[j] = i;
        pri[j] = false;
      }
    }
  }
}
bool calCheck(long long mid) {
  memset(val, 0, sizeof(val));
  long long t;
  for (int i = 0; i < tot; i++) {
    t = prim[i];
    while (mid / t) {
      val[prim[i]] += mid / t;
      t *= prim[i];
    }
  }
  for (int i = 0; i < tot; i++) {
    if (num[prim[i]] > val[prim[i]]) return false;
  }
  return true;
}
void cal() {
  for (int i = MAXM - 1; i >= 2; i--) {
    if (num[i]) num[i - 1] += num[i];
  }
  for (int i = MAXM - 1; i >= 2; i--) {
    if (!pri[i]) {
      num[i / minP[i]] += num[i];
      num[minP[i]] += num[i];
    }
  }
  long long ll = 0, rr = sum;
  while (ll + 1 < rr) {
    long long mid = (ll + rr) / 2;
    if (calCheck(mid)) {
      rr = mid;
    } else {
      ll = mid;
    }
  }
  cout << rr << endl;
}
int main() {
  getPrim();
  scanf("%d", &k);
  sum = 0;
  for (int i = 0; i < k; i++) {
    int tmp;
    scanf("%d", &tmp);
    sum += tmp;
    num[tmp]++;
  }
  cal();
  return 0;
}