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 + 26;
const int MO = 1e9 + 7;
int n, a[N], b[N], c[N], d[N], e[N], r[N];
int main() {
  int i, j, x, f;
  scanf("%d", &n);
  while (n--) {
    scanf("%d", &x);
    a[x]++;
  }
  n = 2e6;
  for (i = 2; i <= n; i = i + 1)
    for (j = i + i; j <= n; j = j + i) e[j] = 1;
  for (i = n; i > 1; i = i - 1) {
    if (!a[i]) continue;
    a[i]--;
    x = 0;
    for (j = i; j <= n; j = j + i) x |= b[j];
    if (x)
      b[i - 1]++;
    else
      b[i]++;
    i++;
  }
  for (i = 1; i <= n; i = i + 1)
    for (j = i; j <= n; j = j + i)
      if (b[j]) c[i] += b[j], d[i] = j;
  f = 1;
  for (i = 2; i <= n; i = i + 1) {
    if (e[i]) continue;
    x = 1;
    while ((long long)x * i <= n && c[x * i]) {
      x *= i;
    }
    f = (long long)f * x % MO;
    if (c[x] == 1) r[d[x]] = 1;
  }
  for (i = 1; i <= n; i = i + 1) {
    if (b[i] && !r[i]) {
      f++;
      break;
    }
  }
  cout << f % MO;
  return 0;
}