#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 7, M = 1e9 + 7;
bool isp[N];
vector<pair<int, int> > factorize(int x) {
  vector<pair<int, int> > ans;
  for (int i = 2; i * i <= x; ++i) {
    int cnt = 0;
    while (x % i == 0) cnt++, x /= i;
    if (cnt) ans.push_back(make_pair(i, cnt));
  }
  if (x > 1) ans.push_back(make_pair(x, 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int pp[n];
  for (int i = 0; i < n; i++) cin >> pp[i];
  sort(pp, pp + n);
  reverse(pp, pp + n);
  long long ans = 1;
  map<int, int> fac;
  map<int, int> cnt;
  vector<vector<pair<int, int> > > factors;
  for (int i = 1; i <= n; i++) {
    int p = pp[i - 1];
    if (fac.count(p) == 0) {
      ans = (ans * p) % M;
      fac[p] = 1;
      cnt[p] = 1;
      vector<pair<int, int> > v;
      v.push_back(make_pair(p, 1));
      factors.push_back(v);
    } else {
      vector<pair<int, int> > pf = factorize(p - 1);
      factors.push_back(pf);
      for (auto pr : pf) {
        int p = pr.first;
        int e = pr.second;
        int dif = e - fac[p];
        if (dif == 0)
          cnt[p]++;
        else if (dif > 0) {
          fac[p] += dif;
          cnt[p] = 1;
          for (int i = 0; i < dif; i++) ans = (ans * p) % M;
        }
      }
    }
  }
  for (auto pf : factors) {
    bool found = true;
    for (auto pr : pf) {
      int p = pr.first;
      int e = pr.second;
      if (fac[p] == e && cnt[p] == 1) {
        found = false;
        break;
      }
    }
    if (found) {
      ans++;
      break;
    }
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 7, M = 1e9 + 7;
bool isp[N];
vector<pair<int, int> > factorize(int x) {
  vector<pair<int, int> > ans;
  for (int i = 2; i * i <= x; ++i) {
    int cnt = 0;
    while (x % i == 0) cnt++, x /= i;
    if (cnt) ans.push_back(make_pair(i, cnt));
  }
  if (x > 1) ans.push_back(make_pair(x, 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int pp[n];
  for (int i = 0; i < n; i++) cin >> pp[i];
  sort(pp, pp + n);
  reverse(pp, pp + n);
  long long ans = 1;
  map<int, int> fac;
  map<int, int> cnt;
  vector<vector<pair<int, int> > > factors;
  for (int i = 1; i <= n; i++) {
    int p = pp[i - 1];
    if (fac.count(p) == 0) {
      ans = (ans * p) % M;
      fac[p] = 1;
      cnt[p] = 1;
      vector<pair<int, int> > v;
      v.push_back(make_pair(p, 1));
      factors.push_back(v);
    } else {
      vector<pair<int, int> > pf = factorize(p - 1);
      factors.push_back(pf);
      for (auto pr : pf) {
        int p = pr.first;
        int e = pr.second;
        int dif = e - fac[p];
        if (dif == 0)
          cnt[p]++;
        else if (dif > 0) {
          fac[p] += dif;
          cnt[p] = 1;
          for (int i = 0; i < dif; i++) ans = (ans * p) % M;
        }
      }
    }
  }
  for (auto pf : factors) {
    bool found = true;
    for (auto pr : pf) {
      int p = pr.first;
      int e = pr.second;
      if (fac[p] == e && cnt[p] == 1) {
        found = false;
        break;
      }
    }
    if (found) {
      ans++;
      break;
    }
  }
  cout << ans << endl;
}

```