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 = 2000005;
const int mod = 1000000007;
long long p[N];
int max_p[N], cnt[N];
int prime[555555], pcnt;
bool pvis[N];
int n;
void getP(int n) {
  for (int i = 2; i <= n; i++) {
    if (!pvis[i]) {
      prime[++pcnt] = i;
    }
    for (int j = 1; j <= pcnt; j++) {
      if (prime[j] * i > n) break;
      pvis[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
long long qpow(int a, int k) {
  long long res = 1, base = a;
  while (k) {
    if (k & 1) {
      res = (res * base) % mod;
    }
    base = (base * base) % mod;
    k /= 2;
  }
  return res;
}
inline void update(int p, int k) {
  if (k > max_p[p]) {
    max_p[p] = k;
    cnt[p] = 1;
  } else if (k == max_p[p]) {
    ++cnt[p];
  }
}
bool check(int x) {
  if (pvis[x]) {
    for (int i = 1; i <= pcnt; i++) {
      if (prime[i] * prime[i] > x) {
        break;
      }
      int k = 0;
      while (x % prime[i] == 0) {
        x /= prime[i];
        k++;
      }
      if (max_p[prime[i]] == k && cnt[prime[i]] == 1) {
        return false;
      }
    }
    if (x > 1 && max_p[x] == 1 && cnt[x] == 1) return false;
    return true;
  } else {
    if (max_p[x] == 1 && cnt[x] == 1) return false;
    return true;
  }
}
long long lcm = 1;
int nya = 3;
bool vis[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  sort(p + 1, p + n + 1);
  getP(p[n]);
  for (int i = n; i >= 1; i--) {
    if (max_p[p[i]]) {
      int val = p[i] - 1;
      for (int j = 1; j <= pcnt; j++) {
        if (prime[j] * prime[j] > val) break;
        int k = 0;
        while (val % prime[j] == 0) {
          val /= prime[j];
          k++;
        }
        k != 0 ? update(prime[j], k) : (void)nya;
      }
      if (val > 1) {
        update(val, 1);
      }
      vis[i] = 1;
    } else {
      update(p[i], 1);
    }
  }
  for (int i = 1; i <= pcnt; i++) {
    if (max_p[prime[i]]) {
      lcm = 1ll * lcm * qpow(prime[i], max_p[prime[i]]) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    if ((vis[i] && check(p[i] - 1)) || (!vis[i] && check(p[i]))) {
      lcm = (lcm + 1) % mod;
      break;
    }
  }
  cout << lcm;
}