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;
int mx[2000005];
int A[2000005], tp[2000005];
bool use[2000005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%d", &p);
    A[i] = p;
  }
  sort(A, A + n);
  bool up = false;
  for (int i = n - 1; i >= 0; i--) {
    int p = A[i];
    if (mx[p] == 0)
      mx[p] = 1, tp[i] = 0;
    else {
      tp[i] = 1;
      p--;
      bool ok = false;
      for (int j = 2; j * j <= p; j++) {
        if (p % j == 0) {
          int c = 0;
          while (p % j == 0) p /= j, c++;
          if (mx[j] < c) {
            mx[j] = c;
            ok = true;
          }
        }
      }
      if (p != 1 && mx[p] == 0) {
        mx[p] = 1;
        ok = true;
      }
      if (!ok) up = true;
    }
  }
  long long int ret = 1;
  for (int i = 2; i < 2000005; i++)
    for (int j = 0; j < mx[i]; j++) ret = ret * (long long int)i % 1000000007;
  if (!up) {
    for (int i = 0; i < n; i++) {
      int p = A[i];
      if (tp[i] == 0)
        use[p] = true;
      else {
        p--;
        bool ok = false;
        for (int j = 2; j * j <= p; j++) {
          if (p % j == 0) {
            int c = 0;
            while (p % j == 0) p /= j, c++;
            if (mx[j] == c && !use[j]) {
              use[j] = true;
              ok = true;
            }
          }
        }
        if (p != 1 && mx[p] == 1 && !use[p]) {
          use[p] = true;
          ok = true;
        }
        if (!ok) up = true;
      }
    }
  }
  printf("%lld\n", (ret + up) % 1000000007);
  return 0;
}