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 = 2e6 + 10;
const int Mod = 1e9 + 7;
int prm[N], fac[N], tot;
void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (!prm[i]) prm[++tot] = i, fac[i] = i;
    for (int j = 1; j <= tot && i * prm[j] <= n; ++j) {
      int x = i * prm[j];
      prm[x] = true, fac[x] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
}
int n;
int A[N], cnt[N];
bool flag[N], vis[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  init(N - 5);
  sort(A + 1, A + n + 1, greater<int>());
  for (int i = 1; i <= n; i++) {
    if (!cnt[A[i]])
      ++cnt[A[i]], flag[A[i]] = false, vis[i] = true;
    else {
      int x = A[i] - 1;
      while (x > 1) {
        int d = fac[x], c = 0;
        while (x % d == 0) ++c, x /= d;
        if (c > cnt[d])
          cnt[d] = c, flag[d] = false;
        else if (c == cnt[d])
          flag[d] = true;
      }
    }
  }
  int ans = 1;
  for (int i = 1; i <= N - 5; i++)
    for (int j = 1; j <= cnt[i]; j++) ans = 1ll * ans * i % Mod;
  bool ok = false;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      int x = A[i] - 1;
      bool del = true;
      while (x > 1) {
        int d = fac[x], c = 0;
        while (x % d == 0) ++c, x /= d;
        if (c == cnt[d] && !flag[d]) {
          del = false;
          break;
        }
      }
      if (del) {
        ok = true;
        break;
      }
    }
  ans = (ans + ok) % Mod;
  printf("%d\n", ans);
  return 0;
}