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 mo = 1000000007;
const int N = 200005, M = 2000000;
int n, a[N], b[N], cnt[M + 5];
int mn[M + 5], mx[M + 5], t[M + 5];
void init() {
  for (int i = (int)(0); i <= (int)(M + 2); i++) mn[i] = 233333333;
  for (int i = (int)(2); i <= (int)(M); i++)
    for (int j = (int)(1); j <= (int)(M / i); j++)
      mn[i * j] = min(mn[i * j], i);
}
void insert(int v) {
  b[++*b] = v;
  for (; v != 1;) {
    int p = mn[v], cnt = 0;
    for (; v % p == 0; v /= p, cnt++)
      ;
    if (cnt > mx[p])
      mx[p] = cnt, t[p] = 1;
    else if (cnt == mx[p])
      t[p]++;
  }
}
bool check(int v) {
  for (; v != 1;) {
    int p = mn[v], cnt = 0;
    for (; v % p == 0; v /= p, cnt++)
      ;
    if (cnt == mx[p] && t[p] == 1) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]), cnt[a[i]]++;
  for (int i = (int)(1); i <= (int)(M); i++)
    if (cnt[i] >= 2) insert(i), insert(i - 1);
  for (int i = (int)(M); i >= (int)(1); i--)
    if (cnt[i] == 1)
      if (mx[i])
        insert(i - 1);
      else
        insert(i);
  int ans = 1, ans2 = 0;
  for (int i = (int)(1); i <= (int)(M); i++)
    for (int j = (int)(1); j <= (int)(mx[i]); j++) ans = 1ll * ans * i % mo;
  for (int i = (int)(1); i <= (int)(M); i++)
    if (cnt[i] >= 3 - (i == 2)) ans2 = 1;
  for (int i = (int)(1); i <= (int)(*b); i++)
    if (check(b[i])) ans2 = 1;
  printf("%d\n", (ans + ans2) % mo);
}