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 mod = 1e9 + 7;
const int N = 2000010;
int n, cnt[N], mx = 0;
int pri[N], d[N], ds[N];
bool ty[N];
inline int add(const int &x, const int &y) {
  return x + y < mod ? x + y : x + y - mod;
}
inline int sub(const int &x, const int &y) {
  return x - y < 0 ? x - y + mod : x - y;
}
inline int mul(const int &x, const int &y) {
  return (int)((long long)x * y % mod);
}
inline int Max(const int &x, const int &y) { return x > y ? x : y; }
void mdy(int i, int x) {
  x > d[i] ? d[i] = x, ds[i] = 1 : x == d[i] ? ++ds[i] : 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) scanf("%d", &x), mx = Max(mx, x), ++cnt[x];
  for (int i = 2; i <= mx; i++)
    if (!pri[i])
      for (int j = i; j <= mx; j += i)
        if (!pri[j]) pri[j] = i;
  int Circle = 1, Pre = 0;
  for (int i = mx; i; i--)
    for (int j = cnt[i]; j--;) {
      if (d[i]) {
        for (int x = i - 1; x > 1;) {
          int p = pri[x], c = 0;
          for (; x % p == 0; x /= p) ++c;
          for (int e = d[p]; e < c; ++e) Circle = mul(Circle, p);
          mdy(p, c);
        }
        ty[i] = 1;
      } else
        mdy(i, 1), Circle = mul(Circle, i);
    }
  for (int i = mx; i; i--)
    if (ty[i]) {
      bool flag = 1;
      for (int x = i - 1; x > 1;) {
        int p = pri[x], c = 0;
        for (; x % p == 0; x /= p) ++c;
        if (c == d[p] && ds[p] == 1) flag = 0;
      }
      if (flag) Pre = 1;
    }
  printf("%d\n", add(Circle, Pre));
  return 0;
}