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 mod = 1e9 + 7;
int p[200005], low[2000005], mx[2000005], cnt[2000005];
vector<pair<int, int>> v[200005];
void update(int x, int y) {
  if (y > mx[x])
    mx[x] = y, cnt[x] = 1;
  else if (y == mx[x])
    cnt[x]++;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  reverse(p + 1, p + n + 1);
  for (int i = 2; i <= p[1]; i++)
    if (!low[i])
      for (int j = i; j <= p[1]; j += i)
        if (!low[j]) low[j] = i;
  for (int i = 1; i <= n; i++) {
    if ((!mx[p[i]]) || p[i] == 2) {
      update(p[i], 1);
      continue;
    }
    int t = p[i] - 1;
    while (t != 1) {
      if ((!v[i].size()) || v[i].back().first != low[t])
        v[i].push_back({low[t], 1});
      else
        v[i][v[i].size() - 1].second++;
      update(v[i].back().first, v[i].back().second);
      t /= low[t];
    }
  }
  int res = 1;
  for (int i = 1; i <= p[1]; i++) {
    int t = mx[i];
    while (t--) res = 1ll * res * i % mod;
  }
  int tmp = 0;
  for (int i = 1; i <= n; i++) {
    if (!v[i].size()) {
      if (mx[p[i]] > 1 || cnt[p[i]] >= 2) {
        tmp = 1;
        break;
      }
      continue;
    }
    bool flag = 1;
    for (auto j : v[i])
      if (mx[j.first] == j.second && cnt[j.first] < 2) {
        flag = 0;
        break;
      }
    if (flag) tmp = 1;
  }
  printf("%d", (res + tmp) % mod);
  return 0;
}