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;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  char c;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') {
    int x = (c = getchar()) - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return -x;
  }
  int x = c - '0';
  while ((c = getchar()) >= '0') x = x * 10 + c - '0';
  return x;
}
const int N = 2e5 + 5, U = 2e6 + 5, D = 1e9 + 7;
int ap[U];
int mx[U], mx_cnt[U];
int r[U];
int pr[U], pcnt;
void init(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!r[i]) r[i] = pr[++pcnt] = i;
    for (int j = 1, x; x = pr[j], i * x <= n; ++j) {
      r[i * x] = x;
      if (x == r[i]) break;
    }
  }
}
int a[N], top;
void add(int x) {
  a[++top] = x;
  while (x > 1) {
    int p = r[x], y = 1;
    while (x % p == 0) {
      x /= p;
      y *= p;
    }
    if (mx[p] < y) {
      mx[p] = y;
      mx_cnt[p] = 1;
    } else
      mx_cnt[p] += mx[p] == y;
  }
}
bool no_value(int x) {
  while (x > 1) {
    int p = r[x], y = 1;
    while (x % p == 0) {
      x /= p;
      y *= p;
    }
    if (mx[p] == y && mx_cnt[p] == 1) return 0;
  }
  return 1;
}
int main() {
  init(U - 1);
  int n = read();
  for (int i = 1; i <= n; ++i) ++ap[read()];
  for (int i = 1; i <= U - 1; ++i) mx[i] = mx_cnt[i] = 1;
  bool ap3 = 0;
  for (int i = U - 1; i >= 2; --i)
    if (ap[i])
      if (ap[i] >= 2) {
        add(i);
        add(i - 1);
        ap3 += ap[i] >= 3;
      } else
        add(mx[i] > 1 ? i - 1 : i);
  for (int i = 1; i <= n; ++i) ap3 |= no_value(a[i]);
  long long ans = 1;
  for (int i = 1; i <= U - 1; ++i) ans = ans * mx[i] % D;
  cout << (ans + ap3) % D;
}