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 cnt[2000010];
int mark[2000010];
int ans[2000010];
int d[2000010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i < 2000010; i++) {
    if (!d[i])
      for (int j = i; j < 2000010; j += i) {
        d[j] = i;
      }
  }
  int n;
  cin >> n;
  vector<int> v(n);
  int flag = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  vector<int> a;
  for (auto x : v) {
    if (ans[x]) x--;
    a.push_back(x);
    while (x > 1) {
      int p = d[x];
      int q = 0;
      while (x % p == 0) {
        x /= p;
        q++;
      }
      if (ans[p] < q) {
        ans[p] = q;
        cnt[p] = 1;
      } else if (ans[p] == q)
        cnt[p]++;
    }
  }
  memset(mark, 0, sizeof(mark));
  for (auto x : a) {
    int flag1 = 0;
    while (x > 1 && !flag1) {
      int p = d[x];
      int q = 0;
      while (x % p == 0) {
        x /= p;
        q++;
      }
      if (ans[p] == q && cnt[p] == 1) flag1 = 1;
    }
    flag |= !flag1;
  }
  long long res = 1LL;
  for (long long i = 2; i < 2000010; i++) {
    if (ans[i]) {
      for (int j = 0; j < ans[i]; j++) res = (res * i) % 1000000007LL;
    }
  }
  res += flag;
  res %= 1000000007LL;
  cout << res << "\n";
  return 0;
}