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 N = 10000005;
vector<int> vv;
long long cnt[N];
int p[N];
int Max;
long long sum;
int n;
void pre() {
  for (int i = 2; i < N; ++i) {
    if (!p[i]) {
      p[i] = i;
      vv.push_back(i);
    }
    for (int j = 0; i * vv[j] < N; ++j) {
      p[i * vv[j]] = vv[j];
      if (i % vv[j] == 0) break;
    }
  }
  return;
}
long long cal(long long mid, long long base) {
  long long ret = 0;
  while (mid) {
    ret += (mid /= base);
  }
  return ret;
}
void deal() {
  for (int i = Max; i >= 2; --i) {
    if (p[i] != i) {
      cnt[p[i]] += cnt[i];
      cnt[i / p[i]] += cnt[i];
    }
  }
}
long long get(long long base, long long x) {
  long long l = 1, r = sum;
  long long ret;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (cal(mid, base) >= x) {
      ret = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  return ret;
}
int main() {
  pre();
  while (scanf("%d", &n) != EOF) {
    memset(cnt, 0, sizeof(cnt));
    sum = 0;
    Max = 0;
    int x;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &x);
      sum += x;
      Max = max(Max, x);
      ++cnt[x];
    }
    for (int i = Max - 1; i >= 2; --i) {
      cnt[i] += cnt[i + 1];
    }
    deal();
    long long ret = 1;
    for (int i = 0; i < vv.size(); ++i) {
      ret = max(ret, get(vv[i], cnt[vv[i]]));
    }
    printf("%I64d\n", ret);
  }
  return 0;
}