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;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
const int MO = 1e9 + 7;
bool u[2000010];
int n, a[2000010], cnt[2000010], p[2000010], f[2000010], pr, P[2000010], t,
    ans[2000010];
long long ANS;
void getp() {
  n = 2e6;
  for (int i = 2; i <= n; i++) {
    if (!u[i]) {
      p[++pr] = i;
      P[i] = i;
      f[i] = i;
    }
    for (int j = 1; j <= pr; j++) {
      if (i * p[j] > n) break;
      u[i * p[j]] = 1;
      if (i % p[j] == 0) {
        f[i * p[j]] = f[i] * p[j];
        P[i * p[j]] = p[j];
        break;
      }
      f[i * p[j]] = p[j];
      P[i * p[j]] = p[j];
    }
  }
}
int main() {
  getp();
  cin >> n;
  for (int i = 1; i <= n; i++) getint(a[i]);
  sort(a + 1, a + n + 1);
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (ans[a[i]]) --a[i];
    int t = a[i];
    while (t > 1) {
      if (f[t] > ans[P[t]]) {
        ans[P[t]] = f[t];
        cnt[P[t]] = 1;
      } else if (f[t] == ans[P[t]]) {
        ++cnt[P[t]];
      }
      t /= f[t];
    }
  }
  ANS = 1;
  for (int i = 1; i <= 2000000; i++)
    if (ans[i]) ANS = ANS * ans[i] % MO;
  for (int i = 1; i <= n; i++) {
    int t = a[i];
    bool ok = 1;
    while (t > 1) {
      if (ans[P[t]] == f[t] && cnt[P[t]] == 1) ok = 0;
      t /= f[t];
    }
    if (ok) {
      (++ANS) %= MO;
      break;
    }
  }
  cout << ANS << endl;
  return 0;
}