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 int mxN = 2e5, mxA = 2e6, M = 1e9 + 7;
int n, p[mxN], spf[mxA + 1];
vector<int> pfs;
array<int, 2> a[mxA + 1];
bool c[mxN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i <= mxA; ++i) {
    if (!spf[i]) {
      pfs.push_back(i);
      spf[i] = i;
    }
    for (int j = 0; j < pfs.size() && pfs[j] <= spf[i] && i * pfs[j] <= mxA;
         ++j)
      spf[i * pfs[j]] = pfs[j];
  }
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i];
  sort(p, p + n, greater<int>());
  for (int i = 0; i < n; ++i) {
    c[i] = a[p[i]][0];
    int b = c[i] ? p[i] - 1 : p[i];
    while (spf[b]) {
      int c = spf[b], d = 0;
      while (b % c == 0) {
        b /= c;
        ++d;
      }
      if (a[c][0] < d) a[c] = {d};
      if (a[c][0] == d) ++a[c][1];
    }
  }
  long long ans = 1;
  for (int i = 2; i <= mxA; ++i)
    for (int j = 0; j < a[i][0]; ++j) ans = i * ans % M;
  for (int i = 0; i < n; ++i) {
    if (!c[i]) continue;
    int b = p[i] - 1;
    bool ok = 1;
    while (spf[b] && ok) {
      int c = spf[b], d = 0;
      while (b % c == 0) {
        b /= c;
        ++d;
      }
      ok = a[c] > array<int, 2>{d, 1};
    }
    if (ok) {
      ++ans;
      break;
    }
  }
  cout << ans;
}