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 = 2e6 + 10;
const int mod = 1e9 + 7;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, pri[N], tt, minp[N], mx[N], op[N], cnt[N], p[N];
bool vis[N];
vector<pair<int, int> > fac(int x) {
  vector<pair<int, int> > ret;
  while (x > 1) {
    int y = minp[x];
    ret.push_back(make_pair(y, 0));
    while (x % y == 0) ++ret.back().second, x /= y;
  }
  return ret;
}
int main() {
  for (int i = 2; i < N; i++)
    if (!vis[i]) {
      pri[++tt] = i;
      minp[i] = i;
      for (int j = i + i; j < N; j += i) {
        vis[j] = 1;
        if (!minp[j]) minp[j] = i;
      }
    }
  cin >> n;
  for (int i = 1; i <= n; i++) p[i] = gi();
  sort(p + 1, p + n + 1, greater<int>());
  for (int i = 1; i <= n; i++) {
    if (!mx[p[i]])
      mx[p[i]] = cnt[p[i]] = 1, op[i] = 1;
    else {
      auto vec = fac(p[i] - 1);
      for (auto x : vec) {
        if (mx[x.first] < x.second)
          mx[x.first] = x.second, cnt[x.first] = 1;
        else if (mx[x.first] == x.second)
          ++cnt[x.first];
      }
    }
  }
  int ans = 1;
  bool fl = 0;
  for (int i = 1; i < N; i++) {
    for (int j = 1; j <= mx[i]; j++) ans = 1ll * ans * i % mod;
  }
  for (int i = 1; i <= n && !fl; i++)
    if (op[i]) {
      fl |= (mx[p[i]] > 1 || cnt[p[i]] > 1);
    } else {
      auto vec = fac(p[i] - 1);
      bool all = 1;
      for (auto x : vec) all &= (mx[x.first] > x.second || cnt[x.first] > 1);
      fl |= all;
    }
  if (fl) ans = (ans + 1) % mod;
  cout << ans;
  return 0;
}