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;
namespace RenaMoe {
const int N = 2e6 + 5;
const int P = 1e9 + 7;
inline int power(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % P;
    x = (long long)x * x % P, y >>= 1;
  }
  return res;
}
int n, cntp, ans;
int a[N], pri[N], ma[N], cnt[N];
bool np[N], wc[N];
inline void get_prime(int lim) {
  np[1] = true;
  for (int i = 2; i <= lim; ++i) {
    if (!np[i]) pri[++cntp] = i;
    for (int j = 1; j <= cntp && pri[j] * i <= lim; ++j) {
      np[pri[j] * i] = true;
      if (i % pri[j] == 0) break;
    }
  }
}
inline void update(int p, int c) {
  if (c > ma[p])
    ma[p] = c, cnt[p] = 1;
  else if (c == ma[p])
    ++cnt[p];
}
inline void deco(int x) {
  for (int i = 1; i <= cntp && pri[i] * pri[i] <= x; ++i) {
    int p = pri[i];
    if (x % p) continue;
    int c = 0;
    while (x % p == 0) x /= p, ++c;
    update(p, c);
  }
  if (x > 1) update(x, 1);
}
inline bool check(int x) {
  if (!np[x]) {
    if (ma[x] == 1 && cnt[x] == 1) return false;
    return true;
  }
  for (int i = 1; i <= cntp && pri[i] * pri[i] <= x; ++i) {
    int p = pri[i];
    if (x % p) continue;
    int c = 0;
    while (x % p == 0) x /= p, ++c;
    if (ma[p] == c && cnt[p] == 1) return false;
  }
  if (x > 1 && ma[x] == 1 && cnt[x] == 1) return false;
  return true;
}
inline void main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1), reverse(a + 1, a + n + 1);
  get_prime(a[1]);
  ans = 1;
  for (int i = 1; i <= n; ++i) {
    if (ma[a[i]])
      deco(a[i] - 1), wc[i] = true;
    else
      update(a[i], 1);
  }
  for (int i = 1; i <= cntp; ++i) {
    if (!ma[pri[i]]) continue;
    ans = (long long)ans * power(pri[i], ma[pri[i]]) % P;
  }
  for (int i = 1; i <= n; ++i) {
    if ((wc[i] && check(a[i] - 1)) || (!wc[i] && check(a[i]))) {
      ++ans;
      break;
    }
  }
  cout << ans << '\n';
}
}  // namespace RenaMoe
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  RenaMoe::main();
  cout.flush();
  return 0;
}