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 M = 1e7 + 10;
int pre[M];
long long cnt[M];
vector<int> pv;
void getpre() {
  for (int i = 2; i < M; i++) {
    if (!pre[i]) {
      pre[i] = i;
      pv.push_back(i);
    }
    int len = pv.size();
    for (int j = 0; j < len && i * pv[j] < M; j++) {
      pre[i * pv[j]] = pv[j];
      if (i % pv[j] == 0) break;
    }
  }
}
long long cal(long long m, int d) {
  long long ans = 0;
  while (m >= d) {
    ans = ans + m / d;
    m = m / d;
  }
  return ans;
}
long long midfind(int d, long long sum) {
  long long low = 1, high = sum, mid, ans = sum;
  while (low <= high) {
    mid = (high + low) >> 1;
    if (cal(mid, d) >= cnt[d]) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  return ans;
}
int main() {
  int k, a, Max;
  getpre();
  while (cin >> k) {
    memset(cnt, 0, sizeof(cnt));
    Max = 0;
    long long sum = 0;
    for (int i = 0; i < k; i++) {
      scanf("%d", &a);
      cnt[a]++;
      sum += a;
      Max = max(Max, a);
    }
    for (int i = Max - 1; i >= 2; i--) {
      cnt[i] += cnt[i + 1];
    }
    for (int i = Max; i >= 2; i--) {
      if (pre[i] != i) {
        cnt[pre[i]] += cnt[i];
        cnt[i / pre[i]] += cnt[i];
      }
    }
    int len = pv.size();
    long long ans = 1;
    for (int i = 0; i < len; i++) {
      ans = max(ans, midfind(pv[i], sum));
    }
    printf("%I64d\n", ans);
  }
  return 0;
}