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;
long long Pr[5000010], A[5000010], lw[5000010], S[5000010], O[5000010],
    SS[5000010], t = 0;
bool npr[5000010];
signed main() {
  for (long long i = 2; i <= 5000000; i++) {
    if (!npr[i]) Pr[++t] = i, lw[i] = i;
    for (long long j = 1; Pr[j] * i <= 5000000; j++) {
      npr[Pr[j] * i] = 1;
      lw[Pr[j] * i] = Pr[j];
      if (i % Pr[j] == 0) {
        break;
      }
    }
  }
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    long long a;
    scanf("%lld", &a);
    O[a]++;
  }
  for (long long i = 5000000; i >= 1; i--) {
    if (O[i] && !S[i]) S[i] = 1, SS[i] = 1, O[i]--;
    if (O[i]) {
      long long tmp = i - 1;
      while (tmp != 1) {
        long long w = lw[tmp];
        long long cnt = 0;
        while (tmp % w == 0) tmp /= w, cnt++;
        if (S[w] == cnt)
          SS[w]++;
        else if (cnt > S[w])
          SS[w] = 1;
        S[w] = max(S[w], cnt);
      }
    }
  }
  long long ans = 1;
  for (long long i = 1; i <= 5000000; i++)
    for (long long j = 1; j <= S[i]; j++) ans = ans * i % 1000000007;
  long long s = 0;
  for (long long i = 5000000; i >= 1; i--) {
    long long flag = 1;
    if (O[i] > 1)
      flag = 0;
    else if (O[i]) {
      flag = 0;
      long long tmp = i - 1;
      while (tmp != 1) {
        long long w = lw[tmp];
        long long cnt = 0;
        while (tmp % w == 0) tmp /= w, cnt++;
        if (S[w] == cnt && SS[w] == 1) flag = 1;
      }
    }
    s |= !flag;
  }
  printf("%lld\n", ((ans + s) % 1000000007 + 1000000007) % 1000000007);
  return 0;
}