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>
namespace io {
int F() {
  int F = 1, n = 0;
  char ch;
  while ((ch = getchar()) != '-' && (ch < '0' || ch > '9'))
    ;
  ch == '-' ? F = 0 : n = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') n = n * 10 + ch - '0';
  return F ? n : -n;
}
long long G() {
  long long F = 1, n = 0;
  char ch;
  while ((ch = getchar()) != '-' && (ch < '0' || ch > '9'))
    ;
  ch == '-' ? F = 0 : n = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') n = n * 10 + ch - '0';
  return F ? n : -n;
}
}  // namespace io
int R(int l, int r) { return (rand() << 15 | rand()) % (r - l + 1) + l; }
const int M = 1000000007;
int a[2222222];
int d[2222222];
int p[2222222], pp;
int v[2222222];
int cnt[2222222];
int us[2222222];
int boom, fl;
void dfs(int x, int k = 0) {
  if (x == 1) return;
  int y = x, D = d[x];
  while (y % D == 0) y /= D;
  if ((v[D] < (x / y) ? v[D] = (x / y), 1 : 0))
    cnt[D] = 1;
  else if (v[D] == x / y) {
    if (k == 0) ++cnt[D];
    if (k == 1 && cnt[D] == 1) fl = 1;
  }
  dfs(y, k);
}
int main() {
  int n = io::F();
  for (register int i = 2; i <= 2000000; ++i) {
    if (!d[i]) p[++pp] = i, d[i] = i;
    for (register int j = 1,
                      u = ((2000000 / i) < (d[i]) ? (2000000 / i) : (d[i]));
         j <= pp && p[j] <= u; ++j)
      d[i * p[j]] = p[j];
  }
  for (register int i = 1; i <= n; ++i) a[io::F()]++;
  v[1] = 1;
  for (register int i = 2000000; i >= 2; --i) {
    if (a[i] > 2) boom = 1;
    if (a[i] >= 2) {
      if (v[i] == 0) v[i] = i;
      dfs(i - 1), us[i] = 1;
    }
    if (a[i] == 1) {
      if (!v[i])
        v[i] = i;
      else
        dfs(i - 1), us[i] = 1;
    }
  }
  int ans = 1;
  for (register int i = 2; i <= 2000000; ++i)
    if (v[i]) ans = (long long)ans * v[i] % M;
  for (register int i = 2; i <= 2000000; ++i)
    if (us[i]) fl = 0, dfs(i - 1, 1), fl == 0 ? boom = 1 : 0;
  printf("%d\n", (ans + boom) % M);
  return 0;
}