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 long long mod = 1e9 + 7;
const int maxn = 2e5 + 79, maxp = 2e6 + 79;
vector<int> p(maxn), lp(maxp, -1), pr, num(maxn);
struct divisor {
  int maxi, cnt;
  divisor() : maxi(0), cnt(0) {}
};
void update(divisor& d, int pw) {
  if (pw > d.maxi) d.maxi = pw, d.cnt = 0;
  if (pw == d.maxi) d.cnt++;
}
map<int, divisor> lcm;
vector<pair<int, int> > factorize(int x) {
  vector<pair<int, int> > ans;
  while (x > 1) {
    ans.push_back({lp[x], 0});
    while (x % ans.back().first == 0)
      x /= ans.back().first, ans.back().second++;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < maxp; i++) {
    if (lp[i] == -1) lp[i] = i, pr.push_back(i);
    for (int j = 0; j < pr.size() && pr[j] <= lp[i] && i * pr[j] < maxp; j++)
      lp[i * pr[j]] = pr[j];
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];
  sort(p.begin(), p.end(), greater<int>());
  for (int i = 0; i < n; i++) {
    num[i] = p[i];
    if (lcm.count(p[i])) num[i]--;
    vector<pair<int, int> > v = factorize(num[i]);
    for (pair<int, int> j : v) update(lcm[j.first], j.second);
  }
  int pp = 0;
  for (int i = 0; i < n; i++) {
    vector<pair<int, int> > v = factorize(num[i]);
    bool rmv = true;
    for (pair<int, int> j : v)
      rmv &= ((lcm[j.first].maxi > j.second) || (lcm[j.first].cnt > 1));
    if (rmv) {
      pp++;
      break;
    }
  }
  long long ans = 1;
  for (const auto& i : lcm)
    for (int j = 0; j < i.second.maxi; j++) ans = (ans * i.first) % mod;
  ans = (ans + pp) % mod;
  cout << ans << "\n";
  return 0;
}