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 N = 2000100;
const int mod = 1e9 + 7;
bool isnotprime[N];
int prime[N], cnt, f[N], p[N], c[N], cc[N];
void get_prime() {
  isnotprime[1] = true;
  for (int i = 2; i <= 2e6; i++) {
    if (!isnotprime[i]) prime[++cnt] = i, f[i] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= 2e6; j++) {
      isnotprime[i * prime[j]] = true, f[i * prime[j]] = prime[j];
      if (!(i % prime[j])) break;
    }
  }
}
int main() {
  get_prime();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + 1 + n);
  for (int i = n; i; i--) {
    if (!c[p[i]])
      c[p[i]] = 1, cc[p[i]] = 1;
    else {
      int tmp = --p[i];
      while (tmp != 1) {
        int tp = f[tmp], tc = 0;
        while (!(tmp % tp)) tmp /= tp, tc++;
        if (tc > c[tp])
          c[tp] = tc, cc[tp] = 1;
        else if (tc == c[tp])
          cc[tp]++;
      }
    }
  }
  int ans = 1;
  for (int i = 1; i <= 2e6; i++)
    for (int j = 1; j <= c[i]; j++) ans = 1ll * ans * i % mod;
  for (int i = 1; i <= n; i++)
    if (isnotprime[p[i]]) {
      bool flag = false;
      while (p[i] != 1) {
        int tp = f[p[i]], tc = 0;
        while (!(p[i] % tp)) p[i] /= tp, tc++;
        if (tc == c[tp] && cc[tp] == 1) {
          flag = true;
          break;
        }
      }
      if (!flag) {
        ans = (ans + 1) % mod;
        break;
      }
    }
  printf("%d\n", ans);
  return 0;
}