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 long long M = 1e9 + 7;
const int maxn = 2002002;
int fp[maxn];
int n;
int p[maxn];
int a[maxn];
int hav[maxn];
int hac[maxn];
long long ans = 1;
bool rizadd(int x) {
  bool good = false;
  while (x != 1) {
    int cnt = 0;
    int px = fp[x];
    while (x % px == 0) {
      cnt++;
      x /= px;
    }
    if (hav[px] < cnt) {
      for (; hav[px] < cnt; hav[px]++) {
        ans = ans * px % M;
        good = true;
      }
      hac[px] = 1;
    } else if (hav[px] == cnt) {
      hac[px]++;
    }
  }
  return good;
}
bool add(int i, int x) {
  if (!hav[x]) {
    ans = ans * x % M;
    hav[x] = 1;
    a[i] = x;
    hac[x] = 1;
    return true;
  } else {
    a[i] = x - 1;
    return rizadd(x - 1);
  }
}
bool check(int x) {
  while (x != 1) {
    int cnt = 0;
    int px = fp[x];
    while (x % px == 0) {
      cnt++;
      x /= px;
    }
    if (cnt == hav[px] and hac[px] <= 1) return false;
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i < maxn; i++)
    if (!fp[i])
      for (int j = i; j < maxn; j += i) fp[j] = i;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];
  int rem = 0;
  sort(p, p + n);
  reverse(p, p + n);
  for (int i = 0; i < n; i++)
    if (!add(i, p[i])) rem = 1;
  for (int i = 0; i < n; i++)
    if (check(a[i])) rem = 1;
  cout << (ans + rem) % M << endl;
  return 0;
}