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;
int const N = 200000, M = 2000000, md = 1e9 + 7;
int dv[M + 1], x[N];
pair<int, int> fr[M + 1];
void up(int i, int z) {
  if (z > fr[i].first)
    fr[i] = make_pair(z, 1);
  else if (z == fr[i].first)
    ++fr[i].second;
}
void ad(int x) {
  int lst = 0, z = 0;
  while (x > 1) {
    int d = dv[x];
    if (d != lst) {
      up(lst, z);
      lst = d, z = 1;
    } else
      ++z;
    x /= d;
  }
  up(lst, z);
}
bool useless(int x) {
  int lst = 0, z = 0;
  while (x > 1) {
    int d = dv[x];
    if (d != lst) {
      if (z == fr[lst].first && fr[lst].second == 1) return false;
      lst = d, z = 1;
    } else
      ++z;
    x /= d;
  }
  if (z == fr[lst].first && fr[lst].second == 1) return false;
  return true;
}
int main() {
  for (int i = 2; i < (int)(M + 1); ++i)
    if (!dv[i]) {
      dv[i] = i;
      for (int j = i + i; j <= M; j += i)
        if (!dv[j]) dv[j] = i;
    }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) scanf("%d", x + i);
  sort(x, x + n);
  reverse(x, x + n);
  for (int i = 0; i < (int)(n); ++i)
    if (!fr[x[i]].first)
      up(x[i], 1);
    else
      ad(x[i] = x[i] - 1);
  bool p1 = false;
  fr[0].second = 2;
  for (int i = 0; i < (int)(n); ++i)
    if (useless(x[i])) {
      p1 = true;
      break;
    }
  int an = 1;
  for (int i = 2; i < (int)(M + 1); ++i)
    while (fr[i].first--) an = (long long)an * i % md;
  if (p1 && ++an == md) an = 0;
  printf("%d\n", an);
}