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(3)
using namespace std;
int n, p[2000005], mx[2000005], cnt[2000005];
int prime[2000005], fact[2000005];
bool is_prime[2000005];
int form[2000005];
void add(int p, int x) {
  if (x > mx[p]) {
    mx[p] = x;
    cnt[p] = 1;
  } else if (x == mx[p])
    cnt[p]++;
}
int sieve(int n) {
  int p = 0;
  for (int i = 0; i <= n; i++) is_prime[i] = true;
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      prime[p++] = i;
      fact[i] = i;
    }
    for (int j = 0; j < p; j++) {
      if (prime[j] * i > n) break;
      is_prime[prime[j] * i] = false;
      fact[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  return p;
}
int pow_mod(int a, int i) {
  int s = 1;
  while (i) {
    if (i & 1) s = 1LL * s * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    i >>= 1;
  }
  return s;
}
int main() {
  scanf("%d", &n);
  int pp = sieve(2000000);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  int ans = 1;
  sort(p, p + n);
  for (int i = n - 1; i >= 0; i--) {
    int x = p[i];
    if (!mx[x]) {
      add(x, 1);
      form[i] = x;
    } else {
      form[i] = x - 1;
      for (x = p[i] - 1; x > 1;) {
        int v = fact[x], c = 0;
        while (x % v == 0) x = x / v, c++;
        add(v, c);
      }
    }
  }
  int s = 0;
  for (int i = n - 1; i >= 0; i--) {
    int x = p[i];
    if (form[i] == x) {
      if (mx[x] > 1 || (mx[x] == 1 && cnt[x] > 1)) {
        s = 1;
        break;
      }
    } else {
      bool f = false;
      for (x = p[i] - 1; x > 1;) {
        int v = fact[x], c = 0;
        while (x % v == 0) x = x / v, c++;
        if (mx[v] == c && cnt[v] == 1) {
          f = true;
          break;
        }
      }
      if (!f) {
        s = 1;
        break;
      }
    }
  }
  for (int i = 0; i < pp; i++)
    ans = 1LL * ans * pow_mod(prime[i], mx[prime[i]]) % 1000000007;
  printf("%d\n", (ans + s) % 1000000007);
  return 0;
}