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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, tot, cnt;
long long ans = 1;
int pri[2000010], a[2000010], t[2000010], k[2000010], mn[2000010], val[2000010],
    num[2000010];
bool tag[2000010], vis[2000010];
void init(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!tag[i]) mn[pri[++tot] = i] = i;
    for (int j = 1; j <= tot; ++j) {
      int k = i * pri[j];
      if (k > n) break;
      tag[k] = true, mn[k] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
void get(int x) {
  cnt = 0;
  int m;
  while (x != 1) {
    t[++cnt] = m = mn[x], k[cnt] = 0;
    while (x % m == 0) x /= m, k[cnt]++;
  }
}
bool check(int x) {
  get(x);
  for (int i = 1; i <= cnt; ++i)
    if (num[t[i]] == 1 && val[t[i]] == k[i]) return false;
  return true;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]);
  sort(a + 1, a + n + 1), init(a[n]);
  for (int i = n; i; --i) {
    if (!val[a[i]])
      val[a[i]] = num[a[i]] = vis[i] = 1, ans = ans * a[i] % 1000000007;
    else {
      get(a[i] - 1);
      for (int j = 1; j <= cnt; ++j) {
        if (k[j] > val[t[j]]) {
          for (int l = val[t[j]] + 1; l <= k[j]; ++l)
            ans = ans * t[j] % 1000000007;
          val[t[j]] = k[j], num[t[j]] = 1;
        } else if (k[j] == val[t[j]])
          num[t[j]]++;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    if (check(a[i] - 1)) {
      ans = (ans + 1) % 1000000007;
      break;
    }
  }
  printf("%lld", ans);
  return 0;
}