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 = 2000009;
int lp[N + 1];
vector<int> pr;
void compute() {
  for (int i = 2; i <= N; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < (int)pr.size() && pr[j] <= lp[i] && i * pr[j] <= N; ++j)
      lp[i * pr[j]] = pr[j];
  }
}
map<int, int> power, freq;
int arr[301009];
bool repeated[301009];
int main() {
  compute();
  int n, v, f;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  sort(arr, arr + n);
  reverse(arr, arr + n);
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    v = arr[i];
    if (power.count(v) == 0) {
      power[v] = 1;
      freq[v] = 1;
    } else {
      repeated[i] = true;
      v--;
      while (v != 1) {
        int k = lp[v], f = 0;
        while (v % k == 0) {
          v /= k;
          f++;
        }
        if (power.count(k) == 0 || power[k] < f) {
          power[k] = f;
          freq[k] = 1;
        } else if (power[k] == f)
          freq[k]++;
      }
    }
  }
  for (auto& k : power) {
    v = k.first, f = k.second;
    while (f > 0) {
      ans = (ans * v) % 1000000007;
      f--;
    }
  }
  bool add = false;
  for (int i = 0; i < n; i++) {
    v = arr[i];
    if (repeated[i] == false) {
      if (power[v] > 1 || freq[v] > 1) {
        add = true;
        break;
      }
    } else {
      v--;
      bool ok = true;
      while (v != 1) {
        int k = lp[v], f = 0;
        while (v % k == 0) {
          v /= k;
          f++;
        }
        if (power[k] > f || (power[k] == f && freq[k] > 1))
          ;
        else
          ok = false;
      }
      if (ok) {
        add = true;
        break;
      }
    }
  }
  printf("%lld\n", (ans + (add ? 1 : 0)) % 1000000007);
  return 0;
}