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].
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using pii = pair<int, int>;
const int V = 2e6 + 5, MOD = 1e9 + 7;
map<int, pii> mp;
vector<int> p, phi, prm;
int n;
int lst[V];
static void get_lst() {
  for (int i = 2; i < V; ++i)
    if (lst[i] == 0)
      for (int j = i; j < V; j += i) lst[j] = i;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  i64 ant;
  int pr, ex;
  get_lst();
  cin >> n;
  p.resize(n);
  for (auto &i : p) cin >> i;
  sort(p.rbegin(), p.rend());
  for (auto i : p) {
    if (mp[i] == pii(0, 0)) {
      mp[i] = pii(0, 1);
      prm.push_back(i);
      continue;
    }
    if (i == 2) {
      mp[i].first = max(mp[i].first, 1);
      prm.push_back(2);
      continue;
    }
    i -= 1;
    phi.push_back(i);
    while (i > 1) {
      pr = lst[i];
      ex = 0;
      while (lst[i] == pr) {
        ex += 1;
        i /= lst[i];
      }
      auto &ref = mp[pr];
      if (ref.second <= ex) {
        ref.first = ref.second;
        ref.second = ex;
      } else if (ref.first < ex)
        ref.first = ex;
    }
  }
  ant = 1;
  for (auto i : mp)
    for (int j = 0; j < i.second.second; ++j) ant = ant * i.first % MOD;
  bool flag = false;
  for (auto i : prm) {
    bool ok = true;
    if (mp[i].second > 0 && mp[i].first > 0) {
      ant += 1;
      flag = true;
      break;
    }
  }
  if (!flag)
    for (auto i : phi) {
      bool ok = true;
      while (i > 1) {
        pr = lst[i];
        ex = 0;
        while (lst[i] == pr) {
          ex += 1;
          i /= lst[i];
        }
        auto &ref = mp[pr];
        ok &= ref.second && ex <= ref.first;
      }
      if (ok) {
        ant += 1;
        break;
      }
    }
  cout << ant << endl;
  return 0;
}
