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 n, p[2000005], f[2000005], pn, flag, d[2000005];
long long a[2000005], c[2000005], b[2000005];
long long pw(long long x, long long y) {
  long long ans = 1;
  for (; y; y >>= 1, x = x * x % 1000000007)
    if (y & 1) ans = ans * x % 1000000007;
  return ans;
}
void work(long long x) {
  for (int i = 1; i <= pn; i++) {
    long long tmp = 0;
    while (x % p[i] == 0) x /= p[i], tmp++;
    if (c[i] == tmp)
      d[i] = 2;
    else if (c[i] < tmp)
      c[i] = tmp, d[i] = 1;
    if (x == 1) break;
    if (f[x] > 0) {
      if (c[f[x]] == 1)
        d[f[x]] = 2;
      else if (c[f[x]] == 0)
        c[f[x]] = 1, d[f[x]] = 1;
      break;
    }
  }
}
bool check(long long x) {
  for (int i = 1; i <= pn; i++) {
    long long tmp = 0;
    while (x % p[i] == 0) x /= p[i], tmp++;
    if (c[i] == tmp && d[i] == 1) return 0;
    if (x == 1) break;
    if (f[x] > 0) {
      if (c[f[x]] == 1 && d[f[x]] == 1) return 0;
      break;
    }
  }
  return 1;
}
int main() {
  flag = 0;
  for (int i = 2; i <= 2000000; i++)
    if (f[i] == 0) {
      p[++pn] = i;
      f[i] = pn;
      for (int j = i + i; j <= 2000000; j += i) f[j] = -1;
    }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = n; i >= 1; i--) {
    if (c[f[a[i]]])
      work(a[i] - 1), b[i] = a[i] - 1;
    else
      work(a[i]), b[i] = a[i];
  }
  for (int i = 1; i <= n && !flag; i++)
    if (check(b[i])) flag = 1;
  long long ans = 1;
  for (int i = 1; i <= pn; i++) ans = ans * pw(p[i], c[i]) % 1000000007;
  printf("%lld\n", (ans + flag) % 1000000007);
}