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 main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  vector<int> used(n, 0);
  vector<vector<pair<int, int> > > pfact(n);
  map<int, int> f;
  map<int, int> ffreq;
  for (int i = 0; i < n; i++) {
    if (f[a[i]] == 0) {
      f[a[i]] = 1;
      used[i] = 1;
      ffreq[a[i]] = 1;
    } else {
      a[i]--;
      for (int p = 2; p * p <= a[i]; p++) {
        if (a[i] % p == 0) {
          int k = 0;
          while (a[i] % p == 0) {
            a[i] /= p;
            k++;
          }
          pfact[i].push_back({p, k});
          if (k > f[p]) {
            f[p] = k;
            ffreq[p] = 0;
          }
          if (k == f[p]) {
            ffreq[p]++;
          }
        }
      }
      if (a[i] > 1) {
        int p = a[i];
        int k = 1;
        pfact[i].push_back({p, k});
        if (k > f[p]) {
          f[p] = k;
          ffreq[p] = 0;
        }
        if (k == f[p]) {
          ffreq[p]++;
        }
      }
    }
  }
  const long long MOD = 1000000007;
  long long num = 1;
  for (auto x : f) {
    for (int j = 0; j < x.second; j++) {
      num = (num * (long long)x.first) % MOD;
    }
  }
  int ok = 0;
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      int can_delete = 1;
      for (pair<int, int> x : pfact[i]) {
        int p = x.first;
        int k = x.second;
        if (f[p] == k && ffreq[p] == 1) {
          can_delete = 0;
        }
      }
      if (can_delete == 1) ok = 1;
    }
  }
  num = (num + ok) % MOD;
  cout << num << '\n';
}