Problem: You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.

One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.

The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).

Output

Print one integer — the maximum number of different tuples modulo 10^9 + 7.

Examples

Input

4
2 3 5 7


Output

210


Input

3
5 3 3


Output

30

Note

In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.

In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 2000005, mod = 1e9 + 7;
int n, p[Maxn], maxi[Maxn], cnt[Maxn];
long long ans = 1;
vector<int> Ve;
void update(int x, int c) {
  if (c == maxi[x])
    cnt[x]++;
  else if (c > maxi[x])
    cnt[x] = 1, maxi[x] = c;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + 1 + n);
  for (int i = n; i >= 1; i--) {
    if (!maxi[p[i]])
      update(p[i], 1);
    else {
      Ve.push_back(p[i] - 1);
      int lim = sqrt(p[i] - 1), tmp = p[i] - 1;
      for (int j = 2; j <= lim; j++) {
        int ct = 0;
        while (tmp % j == 0) ct++, tmp /= j;
        update(j, ct);
      }
      if (tmp != 1) update(tmp, 1);
    }
  }
  for (int i = 1; i <= 2000000; i++)
    ans = ((long long)(ans * pow(i, maxi[i]) + 0.5)) % mod;
  for (auto now : Ve) {
    int lim = sqrt(now), tmp = now;
    for (int j = 2; j <= lim; j++) {
      int ct = 0;
      while (tmp % j == 0) ct++, tmp /= j;
      if (ct && ct == maxi[j] && cnt[j] == 1) goto END;
    }
    if (tmp != 1 && maxi[tmp] == 1 && cnt[tmp] == 1) goto END;
    ans++;
    break;
  END:;
  }
  printf("%lld", ans);
  return 0;
}