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 maxn = 2e5 + 10, MAX = 2e6 + 10, mod = 1e9 + 7;
int par[MAX], cnt[MAX], mx[MAX];
vector<int> vec;
queue<int> q;
int on, ans = 1;
int main() {
  for (int i = 2; i < MAX; i++) {
    if (par[i] == 0)
      for (int j = i; j < MAX; j += i) par[j] = i;
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  for (int i = 2; i < MAX; i++) {
    on |= (cnt[i] > 2);
    if (cnt[i] > 1) {
      q.push(i);
      q.push(i - 1);
      cnt[i] = 0;
    }
  }
  while (int((q).size())) {
    int x = q.front();
    q.pop();
    vec.clear();
    bool added = 0;
    while (x > 1) {
      int cn = 0, p = par[x];
      while (x % p == 0) x /= p, cn++;
      if (mx[p] == 0 && cnt[p] == 1) cnt[p] = 0, q.push(p - 1);
      if (cn > mx[p]) added = 1, mx[p] = cn;
    }
    on |= (!added);
  }
  for (int i = 0; i < MAX; i++)
    if (cnt[i] == 1) mx[i] = 1;
  for (int i = 0; i < MAX; i++)
    for (int j = 0; j < mx[i]; j++) ans = 1ll * ans * i % mod;
  printf("%d\n", on + ans);
  return 0;
}