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 N = 3e6;
const int MOD = 1e9 + 7;
int n, res = 1, mx, ct[N], cs[N], imp[N], ff;
int not_prime[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int a;
    scanf("%d", &a);
    mx = max(mx, a);
    ct[a]++;
  }
  for (int i = mx; i >= 1; --i)
    if (ct[i]) {
      int gg = 0;
      for (int j = i + i; j <= mx; j += i)
        if (cs[j]) gg = 1;
      if (!gg) cs[i] = 1, ct[i]--;
      cs[i - 1] += ct[i];
    }
  for (int i = 2; i <= mx; ++i)
    if (!not_prime[i]) {
      for (int j = i + i; j <= mx; j += i) not_prime[j] = 1;
      int q = 0, c = 0;
      for (int j = i; j <= mx; j += i)
        if (cs[j]) {
          int k = j, p = 0;
          while (k % i == 0) k /= i, p++;
          if (p > q)
            q = p, c = cs[j];
          else if (p == q)
            c += cs[j];
        }
      for (int j = i; j <= mx; j += i)
        if (cs[j]) {
          int k = j, p = 0;
          while (k % i == 0) k /= i, p++;
          if (p == q && c == 1) imp[j] = 1;
        }
      while (q--) res = 1ll * res * i % MOD;
    }
  for (int i = 1; i <= mx; ++i)
    if (cs[i] && !imp[i]) ff = 1;
  res = (res + ff) % MOD;
  printf("%d\n", res);
}