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>
int prime[262144], nprime;
int min_factor[2097152];
void euler_siece() {
  for (int i = 2; i < 2097152; ++i) {
    if (!min_factor[i]) prime[nprime++] = i, min_factor[i] = i;
    for (int j = 0; i * prime[j] < 2097152; ++j) {
      min_factor[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  0;
}
int pow(int x, int n, int m) {
  int res = 1;
  while (n) {
    if (n & 1) res = (long long int)res * x % m;
    x = (long long int)x * x % m;
    n >>= 1;
  }
  return res;
}
int a[262144];
int f[2097152], c[2097152];
int main() {
  euler_siece();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  std::sort(a, a + n, std::greater<int>());
  for (int i = 0; i < n; ++i) {
    int k = a[i] -= !!f[a[i]];
    while (k != 1) {
      int d = min_factor[k], cnt = 0;
      while (k % d == 0) k /= d, ++cnt;
      if (f[d] == cnt)
        ++c[d];
      else if (f[d] < cnt)
        f[d] = cnt, c[d] = 1;
    }
  }
  int ans = 1;
  for (int i = 0; i < 2097152; ++i)
    if (f[i])
      ans = (long long int)ans * pow(i, f[i], ((int)1e9 + 7)) % ((int)1e9 + 7);
  for (int i = 0; i < n; ++i) {
    int k = a[i];
    while (k != 1) {
      int d = min_factor[k], cnt = 0;
      while (k % d == 0) k /= d, ++cnt;
      if (cnt == f[d] && c[d] == 1) goto tag_cont;
    }
    ans = (ans + 1) % ((int)1e9 + 7);
    break;
  tag_cont:
    0;
  }
  printf("%d\n", ans);
  return 0;
}