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 = 200010, X = 2000010, P = 1000000007;
int n, pc, ind;
bool vis[X], chose[N];
int a[N], l[X], pr[N], fc[X];
int tag[X];
void apply(int x) {
  int lst = 0, cnt = 0;
  while (x != 1) {
    int pp = fc[x];
    x /= pp;
    if (pp == lst) {
      ++cnt;
    } else {
      if (cnt >= l[lst]) {
        l[lst] = cnt;
        tag[lst] = ind;
      }
      cnt = 1;
      lst = pp;
    }
  }
  if (cnt >= l[lst]) {
    l[lst] = cnt;
    tag[lst] = ind;
  }
}
int main() {
  for (int x = 2; x < X; ++x) {
    if (!vis[x]) {
      fc[x] = x;
      pr[++pc] = x;
    }
    for (int i = 1; x * pr[i] < X; ++i) {
      vis[x * pr[i]] = true;
      fc[x * pr[i]] = pr[i];
      if (x % pr[i] == 0) break;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1, greater<int>());
  for (int i = 1; i <= n; ++i) {
    ind = i;
    if (!l[a[i]])
      apply(a[i]);
    else
      apply(a[i] - 1);
  }
  int ans = 1;
  for (int i = 1; i <= pc; ++i)
    while (l[pr[i]]--) ans = ans * (long long)pr[i] % P;
  for (int x = 1; x < X; ++x) chose[tag[x]] = true;
  ans = (ans + (count(chose + 1, chose + n + 1, true) != n)) % P;
  printf("%d\n", ans);
  return 0;
}