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 maxn = 200010, V = 2000001, P = 1000000007;
int n, ans = 1, a[maxn], ok[maxn];
int mn[V], e[V], nxt[V], cur[V], cnt[V];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 2; i < V; i++)
    if (!mn[i]) {
      for (int j = i; j < V; j += i) mn[j] = i;
    }
  for (int i = 2; i < V; i++) {
    nxt[i] = i;
    while (!(nxt[i] % mn[i])) e[i]++, nxt[i] /= mn[i];
  }
  sort(a + 1, a + n + 1, greater<int>());
  bool flag = 0;
  for (int i = 1; i <= n; i++) {
    if (!cur[a[i]]) {
      cur[a[i]] = cnt[a[i]] = ok[i] = 1;
    } else {
      for (int j = a[i] - 1; j > 1; j = nxt[j]) {
        if (cur[mn[j]] < e[j])
          cur[mn[j]] = e[j], cnt[mn[j]] = 1;
        else if (cur[mn[j]] == e[j])
          cnt[mn[j]]++;
      }
    }
  }
  for (int i = 2; i < V; i++) {
    for (int j = 0; j < cur[i]; j++) ans = 1LL * i * ans % P;
  }
  auto test = [&](int p, int e) { return cnt[p] > 1 || cur[p] > e; };
  for (int i = 1; i <= n; i++) {
    if (ok[i]) {
      flag |= test(a[i], 1);
    } else {
      bool all = 1;
      for (int j = a[i] - 1; j > 1; j = nxt[j]) all &= test(mn[j], e[j]);
      flag |= all;
    }
  }
  printf("%d\n", (ans + flag) % P);
  return 0;
}