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 = 2e5 + 10, A = 2e6 + 10, mod = 1e9 + 7;
int p[N], max_pwr[A], ctr[A], idx[A];
vector<int> primes;
bool not_prime[A];
void sieve();
long long pwr(long long x, long long e);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  sieve();
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  sort(p + 1, p + 1 + n);
  long long ans = 1;
  set<int> check;
  for (int i = n; i >= 1; --i) {
    if (!max_pwr[idx[p[i]]]) {
      max_pwr[idx[p[i]]] = 1;
      ctr[idx[p[i]]] = 1;
      ans = (ans * p[i]) % mod;
    } else {
      check.insert(p[i] - 1);
      int temp = p[i] - 1, sq = sqrt(temp);
      for (int j = 0; primes[j] <= sq; ++j) {
        int c = 0;
        while (temp % primes[j] == 0) {
          temp /= primes[j];
          ++c;
        }
        if (c > max_pwr[j]) {
          ans = (ans * pwr(primes[j], c - max_pwr[j])) % mod;
          max_pwr[j] = c;
          ctr[j] = 1;
        } else if (c == max_pwr[j]) {
          ++ctr[j];
        }
      }
      if (temp > 1) {
        if (1 > max_pwr[idx[temp]]) {
          ans = (ans * temp) % mod;
          max_pwr[idx[temp]] = 1;
          ctr[idx[temp]] = 1;
        } else if (1 == max_pwr[idx[temp]]) {
          ++ctr[idx[temp]];
        }
      }
    }
  }
  for (auto &i : check) {
    int temp = i, sq = sqrt(temp);
    bool need = false;
    for (int j = 0; primes[j] <= sq; ++j) {
      int c = 0;
      while (temp % primes[j] == 0) {
        temp /= primes[j];
        ++c;
      }
      if (c == max_pwr[j] and ctr[j] == 1) {
        need = true;
        break;
      }
    }
    if (temp > 1 and ctr[idx[temp]] == 1) {
      need = true;
    }
    if (!need) {
      ans = (ans + 1) % mod;
      break;
    }
  }
  cout << ans << endl;
}
void sieve() {
  for (int i = 2; i < A; ++i) {
    if (!not_prime[i]) {
      primes.push_back(i);
      idx[i] = primes.size() - 1;
      for (int j = i; j < A; j += i) {
        not_prime[j] = true;
      }
    }
  }
}
long long pwr(long long x, long long e) {
  long long ans = 1;
  x %= mod;
  while (e) {
    if (e & 1) {
      ans = (ans * x) % mod;
    }
    x = (x * x) % mod;
    e /= 2;
  }
  return ans;
}