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;
namespace cp_base {
auto init = (ios::sync_with_stdio(false), cin.tie(nullptr));
}
const long long int MOD = 1e9 + 7;
int main() {
  vector<int> primeFactor(2e6 + 1, -1);
  for (int i = 2; i < primeFactor.size(); ++i) {
    if (primeFactor[i] == -1) {
      for (int j = i; j < primeFactor.size(); j += i) primeFactor[j] = i;
    }
  }
  int n;
  cin >> n;
  vector<int> inp(n);
  for (int& i : inp) cin >> i;
  sort(inp.begin(), inp.end(), greater<int>());
  vector<int> pPow(2e6 + 1, 0), contributors(2e6 + 1, 0), choice(inp.size());
  for (int id = 0; id < inp.size(); ++id) {
    int i = inp[id];
    if (pPow[i] == 0) {
      choice[id] = i;
      pPow[i] = 1;
      contributors[i] = 1;
    } else {
      choice[id] = i - 1;
      int x = i - 1;
      while (x > 1) {
        int f = primeFactor[x];
        int pow;
        for (pow = 0; x % f == 0; ++pow) x /= f;
        if (pPow[f] < pow) {
          pPow[f] = pow;
          contributors[f] = 1;
        } else if (pPow[f] == pow) {
          contributors[f]++;
        }
      }
    }
  }
  bool plus1 = false;
  for (int x : choice) {
    bool contributes = false;
    while (x > 1) {
      int f = primeFactor[x];
      int pow;
      for (pow = 0; x % f == 0; ++pow) x /= f;
      if (pPow[f] == pow && contributors[f] == 1) contributes = true;
    }
    if (!contributes) plus1 = true;
  }
  long long int ans = 1;
  for (int x = 0; x < pPow.size(); ++x) {
    for (int i = 0; i < pPow[x]; ++i) ans = (ans * x) % MOD;
  }
  if (plus1) ans = (ans + 1) % MOD;
  cout << ans << endl;
}