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 = 1e6 + 3;
const int V = 1e7 + 3;
const long long inf = 1e18;
int n, a[N], pri[V], mipri[V], tot, mxp = 0, id[V];
bool che[V];
long long cnt[V];
void init() {
  tot = 0;
  memset(che, 0, sizeof che);
  memset(cnt, 0, sizeof cnt);
  for (int i = 2; i < V; i++) {
    if (!che[i]) {
      mipri[i] = i;
      id[i] = tot;
      pri[tot++] = i;
    }
    for (int j = 0; j < tot; j++) {
      if (i * pri[j] >= V) break;
      che[i * pri[j]] = 1;
      mipri[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
long long get(long long x, int p) {
  if (x < p) return 0;
  return x / p + get(x / p, p);
}
bool check(long long x) {
  for (int i = mxp; i >= 0; i--) {
    if (!cnt[pri[i]]) continue;
    if (get(x, pri[i]) < cnt[pri[i]]) return 0;
  }
  return 1;
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  a[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = a[i - 1] + 1; j <= a[i]; j++) {
      int t = j;
      while (t > 1) {
        cnt[mipri[t]] += n - i + 1;
        mxp = max(mxp, id[mipri[t]]);
        t /= mipri[t];
      }
    }
  }
  long long l = 1, r = inf, mid, ans;
  while (l <= r) {
    mid = l + r >> 1;
    if (check(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans;
  return 0;
}