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 = 2000001, MOD = 1e9 + 7;
int lcm[MAXN] = {0}, cnt[MAXN] = {0}, add = 0, pre[MAXN] = {0};
bool add_as[MAXN] = {0};
void go(long long p, int idx) {
  if (not lcm[p]) return ++lcm[p], void();
  add_as[idx] = true;
  bool upd = false;
  long long n = p - 1;
  for (; n != 1;) {
    long long d = pre[n], t = 0;
    while (n % d == 0) ++t, n /= d;
    if (lcm[d] < t)
      lcm[d] = t, cnt[d] = 1, upd = true;
    else if (lcm[d] == t)
      ++cnt[d];
  }
  if (not upd) add = 1;
}
bool check(long long p, int idx) {
  long long n = p - 1;
  bool necessary = false;
  for (; n != 1;) {
    long long d = pre[n], t = 0;
    while (n % d == 0) ++t, n /= d;
    if (lcm[d] == t and cnt[d] == 1) necessary = true;
  }
  if (not necessary) add = 1;
}
long long powmod(long long x, long long k) {
  long long ans = 1;
  for (long long cur = x % MOD; k; k >>= 1, cur = cur * cur % MOD) {
    if (k & 1) ans = cur * ans % MOD;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long> p(n);
  for (int i = 0; i < n; ++i) cin >> p[i];
  for (int p = 2; p < MAXN; ++p)
    for (int n = p; n < MAXN; n += p)
      if (not pre[n]) {
        pre[n] = p;
      }
  sort(p.rbegin(), p.rend());
  for (int i = 0; i < n; ++i) go(p[i], i);
  for (int i = 0; i < n; ++i)
    if (add_as[i]) check(p[i], i);
  long long ans = 1;
  for (int i = 1; i < MAXN; ++i) (ans *= powmod(i, lcm[i])) %= MOD;
  (ans += add) %= MOD;
  cout << ans << '\n';
}