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 MOD = 1e9 + 7;
const int P_MAX = 2e6 + 5;
vector<int> smallest_factor;
vector<bool> prime;
void sieve(int maximum) {
  assert(maximum > 0);
  smallest_factor.assign(maximum + 1, 0);
  prime.assign(maximum + 1, true);
  prime[0] = prime[1] = false;
  for (int p = 2; p <= maximum; p += 2) {
    prime[p] = p == 2;
    smallest_factor[p] = 2;
  }
  for (int p = 3; p * p <= maximum; p += 2)
    if (prime[p])
      for (int i = p * p; i <= maximum; i += 2 * p)
        if (prime[i]) {
          prime[i] = false;
          smallest_factor[i] = p;
        }
  for (int p = 2; p <= maximum; p++)
    if (prime[p]) smallest_factor[p] = p;
}
int N;
vector<int> P;
vector<int> powers(P_MAX, 0);
vector<int> reach_count(P_MAX, 0);
bool add(int number, bool real = true) {
  bool helpful = false;
  while (number > 1) {
    int p = smallest_factor[number];
    int power = 0;
    do {
      number /= p;
      power++;
    } while (number % p == 0);
    if (!real) {
      helpful = helpful || power > powers[p];
      continue;
    }
    if (power > powers[p]) {
      powers[p] = power;
      helpful = true;
      reach_count[p] = 1;
    } else if (power == powers[p]) {
      reach_count[p]++;
    }
  }
  return helpful;
}
bool removable(int number) {
  bool ok = true;
  while (number > 1) {
    int p = smallest_factor[number];
    int power = 0;
    do {
      number /= p;
      power++;
    } while (number % p == 0);
    if (power == powers[p] && reach_count[p] == 1) ok = false;
  }
  return ok;
}
int main() {
  sieve(P_MAX);
  scanf("%d", &N);
  P.resize(N);
  map<int, int> P_freq;
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
    P_freq[P[i]]++;
  }
  for (auto it = P_freq.rbegin(); it != P_freq.rend(); it++)
    for (int k = 0; k < it->second; k++)
      if (add(it->first, false))
        add(it->first);
      else
        add(it->first - 1);
  bool add_one = false;
  for (auto it = P_freq.rbegin(); it != P_freq.rend(); it++)
    if (it->second >= 2 && removable(it->first - 1)) add_one = true;
  long long answer = 1;
  for (int i = 2; i < P_MAX; i++)
    for (int j = 0; j < powers[i]; j++) answer = answer * i % MOD;
  if (add_one) answer = (answer + 1) % MOD;
  printf("%lld\n", answer);
}