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;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 1000000007;
  if (b & 1) ns = ns * a % 1000000007;
  return ns;
}
int n;
int p[2000005];
int tt[2000005];
bool bpr[1505];
int prs[2000005], prcnt = 0;
int m[2000005][2];
vector<pair<int, int> > c[2000005];
int main() {
  for (int i = 0; i < 1505; i++) bpr[i] = 1;
  for (int j = 2; j < 1505; j++)
    if (bpr[j]) {
      prs[prcnt++] = j;
      for (int k = 2; k * j < 1505; k++) bpr[k * j] = 0;
    }
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  for (int i = n; i >= 1; i--) {
    if (!tt[p[i]])
      c[i].push_back(make_pair(p[i], 1)), tt[p[i]] = 1;
    else {
      int u = p[i] - 1;
      for (int j = 0; j < prcnt; j++) {
        int cur = 0;
        while (u % prs[j] == 0) u /= prs[j], cur++;
        c[i].push_back(make_pair(prs[j], cur));
        tt[prs[j]] = max(tt[prs[j]], cur);
      }
      if (u > 1) c[i].push_back(make_pair(u, 1)), tt[u] = max(tt[u], 1);
    }
  }
  for (int i = 1; i <= n; i++)
    for (auto v : c[i]) {
      int a = v.first, b = v.second;
      if (b > m[a][0])
        m[a][1] = m[a][0], m[a][0] = b;
      else if (b > m[a][1])
        m[a][1] = b;
    }
  int fl = 0;
  for (int i = 1; i <= n; i++) {
    int cs = 1;
    for (auto v : c[i]) {
      int a = v.first, b = v.second;
      if (m[a][0] == b && m[a][0] != m[a][1]) cs = 0;
    }
    if (cs) fl = 1;
  }
  long long ans = 1;
  for (int i = 0; i < 2000005; i++)
    if (tt[i]) ans = ans * ksm(i, tt[i]) % 1000000007;
  if (fl) ans = (ans + 1) % 1000000007;
  cout << ans << endl;
  return 0;
}