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 mod = 1000000007;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
  }
  int mx = *max_element(p.begin(), p.end());
  vector<bool> isprime(mx + 1);
  for (int i = 2; i <= mx; ++i) isprime[i] = true;
  for (int i = 2; i * i <= mx; ++i) {
    if (!isprime[i]) continue;
    for (int j = i * i; j <= mx; j += i) {
      isprime[j] = false;
    }
  }
  vector<int> primes;
  vector<int> pid(mx + 1, -1);
  for (int i = 2; i <= mx; ++i) {
    if (isprime[i]) {
      pid[i] = primes.size();
      primes.push_back(i);
    }
  }
  vector<int> cnt(primes.size()), top1(primes.size()), top2(primes.size());
  for (int i = 0; i < n; ++i) {
    ++cnt[pid[p[i]]];
  }
  vector<bool> pused(primes.size());
  for (int i = primes.size() - 1; i >= 0; --i) {
    if (cnt[i] >= 1 && top1[i] == 0) {
      pused[i] = true;
      top1[i] = 1;
      --cnt[i];
    }
    if (cnt[i] == 0) continue;
    int x = primes[i] - 1;
    for (int j = 0; primes[j] * primes[j] <= x; ++j) {
      int divcnt = 0;
      while (x % primes[j] == 0) {
        x /= primes[j];
        ++divcnt;
      }
      if (divcnt > top1[j]) {
        top2[j] = top1[j];
        top1[j] = divcnt;
      } else if (divcnt > top2[j]) {
        top2[j] = divcnt;
      }
    }
    if (x > 1) {
      if (top1[pid[x]] == 0) {
        top1[pid[x]] = 1;
      } else if (top2[pid[x]] == 0) {
        top2[pid[x]] = 1;
      }
    }
  }
  bool flag = false;
  for (int i = 0; i < primes.size(); ++i) {
    if (cnt[i] == 0) continue;
    if (cnt[i] >= 2) {
      flag = true;
      continue;
    }
    int x = primes[i] - 1;
    bool decrease = false;
    for (int j = 0; primes[j] * primes[j] <= x; ++j) {
      int divcnt = 0;
      while (x % primes[j] == 0) {
        x /= primes[j];
        ++divcnt;
      }
      if (divcnt == top1[j] && top1[j] > top2[j]) {
        decrease = true;
      }
    }
    if (x > 1) {
      if (top1[pid[x]] == 1 && top1[pid[x]] > top2[pid[x]]) {
        decrease = true;
      }
    }
    if (!decrease) flag = true;
  }
  for (int i = 0; i < primes.size(); ++i) {
    if (pused[i] && (top1[i] != 1 || top1[i] == top2[i])) {
      flag = true;
    }
  }
  int ans = 1;
  for (int i = 0; i < primes.size(); ++i) {
    for (int j = 0; j < top1[i]; ++j) {
      ans = (long long)(ans)*primes[i] % mod;
    }
  }
  if (flag) ans = (ans + 1) % mod;
  cout << ans << endl;
  return 0;
}