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;
using INT = long long;
const int mod = 1000000007;
int flag[2001000], cnt[2001000], ans[2001000], pp[2001000], len, vst[2001000];
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
void pre() {
  for (int i = 2; i < 2001000; i++) {
    if (flag[i]) continue;
    pp[++len] = i;
    for (int j = i; j < 2001000; j += i) flag[j] = i;
  }
}
int prime[2001000];
int a[2001000];
int main() {
  pre();
  int n, nn = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    prime[++nn] = p;
  }
  int f = 0;
  sort(prime + 1, prime + nn + 1);
  for (int i = nn; i >= 1; i--) {
    int p = prime[i];
    if (ans[p] == 0) {
      vst[i] = 1;
      ans[p] = 1;
      cnt[p] = 1;
      continue;
    }
    int x = p - 1;
    while (x > 1) {
      int q = flag[x], ct = 0;
      while (x % q == 0) x /= q, ct++;
      if (ct > ans[q]) {
        ans[q] = ct;
        cnt[q] = 1;
        continue;
      }
      if (ct == ans[q]) cnt[q]++;
    }
  }
  for (int i = 1; i <= nn; i++) {
    int p = prime[i];
    if (vst[i]) continue;
    int x = p - 1, ok = 0;
    while (x > 1) {
      int q = flag[x], ct = 0;
      while (x % q == 0) x /= q, ct++;
      if (ct == ans[q] && cnt[q] == 1) {
        ok = 1;
        break;
      }
    }
    if (!ok) {
      f = 1;
      break;
    }
  }
  int t = 1;
  for (int i = 1; i <= len; i++)
    t = (INT)t * power(pp[i], ans[pp[i]], mod) % mod;
  t = (t + f) % mod;
  cout << t << endl;
  return 0;
}