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 = 2000005, mod = 1000000007;
bool jdg, tp[N];
int n, cnt[N], a[N], ct[N];
long long ans;
int getint() {
  char ch;
  int f = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  int x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x * f;
}
long long getLL() {
  char ch;
  int f = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  long long x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x * f;
}
void fuck(int x) {
  int lim = sqrt(x);
  for (int i = 2; i <= lim; ++i)
    if (x % i == 0) {
      int c = 0;
      while (x % i == 0) ++c, x /= i;
      if (c > a[i])
        a[i] = c, ct[i] = 1;
      else if (c == a[i])
        ++ct[i];
    }
  if (x > 1) {
    if (!a[x])
      a[x] = 1, ct[x] = 1;
    else if (a[x] == 1)
      ++ct[x];
  }
}
void check(int x) {
  int lim = sqrt(x);
  bool add = 0;
  for (int i = 2; i <= lim; ++i)
    if (x % i == 0) {
      int c = 0;
      while (x % i == 0) ++c, x /= i;
      if (c == a[i] && ct[i] == 1) add = 1;
    }
  if (x > 1 && a[x] == 1 && ct[x] == 1) add = 1;
  if (!add) jdg = 1;
}
long long getmi(long long a, long long x) {
  long long rt = 1;
  while (x) {
    if (x & 1) rt = rt * a % mod;
    a = a * a % mod, x >>= 1;
  }
  return rt;
}
int main() {
  n = getint(), ans = 1;
  for (int i = 1; i <= n; ++i) ++cnt[getint()];
  for (int i = 2; i <= 2000000; ++i)
    if (cnt[i] >= 2) fuck(i), fuck(i - 1), jdg |= cnt[i] > 2;
  for (int i = 2000000; i >= 2; --i)
    if (cnt[i] == 1)
      if (!a[i])
        fuck(i), tp[i] = 1;
      else
        fuck(i - 1);
  for (int i = 2; i <= 2000000; ++i)
    if (cnt[i] >= 2) check(i), check(i - 1);
  for (int i = 2; i <= 2000000; ++i)
    if (cnt[i] == 1)
      if (tp[i])
        check(i);
      else
        check(i - 1);
  for (int i = 2; i <= 2000000; ++i) ans = ans * getmi(i, a[i]) % mod;
  printf("%lld\n", (ans + jdg) % mod);
  return 0;
}