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>
#pragma GCC optimize "-O3"
using namespace std;
const int MX = 2e6 + 10;
const long long MOD = 1e9 + 7;
vector<int> pr;
int bb[MX];
int n;
int a[MX];
int p[210000];
int cnt[MX];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (int i = 0; i < MX; ++i) bb[i] = i;
  for (int i = 2; i < MX; ++i) {
    if (bb[i] == i) pr.push_back(i);
    for (int j : pr) {
      if (bb[i] < j || i * j >= MX) break;
      bb[i * j] = j;
    }
  }
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i];
  sort(p, p + n);
  reverse(p, p + n);
  int gfl = 0;
  vector<int> vv;
  for (int i = 0; i < n; ++i) {
    if (!a[p[i]])
      ++a[p[i]], ++cnt[p[i]];
    else {
      int x = p[i] - 1;
      vv.push_back(x);
      int fl = 0;
      while (x != 1) {
        int y = bb[x];
        int cc = 0;
        while (x % y == 0) x /= y, ++cc;
        if (a[y] < cc) {
          cnt[y] = 1;
          a[y] = cc;
          fl = 1;
        } else if (a[y] == cc)
          ++cnt[y];
      }
      if (!fl) gfl = 1;
    }
  }
  if (!gfl) {
    for (int x : vv) {
      int fl = 0;
      while (x != 1) {
        int y = bb[x];
        int cc = 0;
        while (x % y == 0) x /= y, ++cc;
        if (a[y] == cc && cnt[y] == 1) {
          fl = 1;
        }
      }
      if (!fl) gfl = 1;
    }
  }
  long long ans = 1;
  for (int i = 0; i < MX; ++i)
    for (int j = 0; j < a[i]; ++j) ans = (ans * i) % MOD;
  if (gfl) ans = (ans + 1) % MOD;
  cout << ans << "\n";
  return 0;
}