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 + 1;
const long long M = 1e9 + 7;
int n, a[N], f[N], cnt[N];
vector<int> vc;
vector<pair<int, int> > fac[N];
bool used[N], vis[N];
void init() {
  for (int i = 2; i < N; i++) {
    if (vis[i]) continue;
    fac[i].push_back(pair<int, int>(i, 1));
    for (int j = i + i; j < N; j += i) {
      vis[j] = 1;
      if (used[j]) {
        int c = 0, x = j;
        while (x % i == 0) {
          c++;
          x /= i;
        }
        fac[j].push_back(pair<int, int>(i, c));
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    used[a[i] - 1] = 1;
  }
  init();
  sort(a, a + n);
  for (int i = n - 1; i >= 0; i--) {
    int x = a[i];
    if (!f[x])
      f[x] = 1, vc.push_back(x);
    else {
      vc.push_back(x - 1);
      for (int j = 0; j < fac[x - 1].size(); j++) {
        pair<int, int> p = fac[x - 1][j];
        int y = p.first;
        f[y] = max(f[y], p.second);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int x = vc[i];
    for (int j = 0; j < fac[x].size(); j++) {
      pair<int, int> p = fac[x][j];
      int y = p.first;
      if (p.second == f[y]) cnt[y]++;
    }
  }
  bool ok = 0;
  for (int i = 0; i < n; i++) {
    int x = vc[i];
    bool g = 1;
    for (int j = 0; j < fac[x].size(); j++) {
      pair<int, int> p = fac[x][j];
      int y = p.first;
      if (p.second == f[y] && cnt[y] == 1) {
        g = 0;
        break;
      }
    }
    if (g) {
      ok = 1;
      break;
    }
  }
  long long ans = 1;
  for (int i = 2; i < N; i++) {
    while (f[i]) {
      ans = ans * i % M;
      f[i]--;
    }
  }
  if (ok) ans++, ans %= M;
  printf("%lld\n", ans);
}