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;
int n, p[2100000], ma[2100000], cnt_ma[2100000];
int mark[2100000];
bool used[2100000], solid[2100000];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  sort(p + 1, p + n + 1);
  for (int i = n; i; i--) {
    if (i < n && p[i] == p[i + 1]) {
      mark[p[i] - 1]++;
    } else {
      bool uu = false;
      for (int j = p[i]; j <= 2000000; j += p[i])
        if (mark[j]) uu = true;
      if (!uu) {
        mark[p[i]]++;
      } else {
        mark[p[i] - 1]++;
      }
    }
  }
  int ans = 1;
  for (int i = 2; i <= 2000000; i++) {
    if (!used[i]) {
      for (int j = 2 * i; j <= 2000000; j += i) used[j] = true;
    } else {
      continue;
    }
    ma[i] = 0;
    for (int j = i; j <= 2000000; j += i)
      if (mark[j]) {
        int cnt = 0;
        int copy = j;
        while (copy % i == 0) {
          copy /= i;
          cnt++;
        }
        if (cnt > ma[i]) {
          ma[i] = cnt;
          cnt_ma[i] = 1;
        } else if (cnt == ma[i]) {
          cnt_ma[i]++;
        }
      }
    for (int j = i; j <= 2000000; j += i)
      if (mark[j]) {
        int cnt = 0;
        int copy = j;
        while (copy % i == 0) {
          copy /= i;
          cnt++;
        }
        if (cnt == ma[i] && cnt_ma[i] == 1) {
          solid[j] = true;
        }
      }
    for (int j = 1; j <= ma[i]; j++) {
      ans = 1LL * ans * i % 1000000007;
    }
  }
  for (int i = 1; i <= 2000000; i++)
    if ((mark[i] > 0 && !solid[i]) || mark[i] > 1) {
      ans++;
      break;
    }
  printf("%d\n", ans % 1000000007);
}