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 N = 2e6 + 10;
const int mod = 1e9 + 7;
int prime[N], pp, pri[N], ans = 1, p[N], n, cnt[N], mx[N];
bool iscomp[N];
void primetable() {
  for (int i = 2; i < N; i++) {
    if (!iscomp[i]) prime[pp++] = i, pri[i] = i;
    for (int j = 0; j < pp && 1ll * i * prime[j] < N; j++) {
      iscomp[i * prime[j]] = 1;
      pri[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  primetable();
  scanf("%d\n", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  for (int i = n; i; i--) {
    if (!cnt[p[i]])
      cnt[p[i]]++, mx[p[i]] = 1, ans = (1ll * ans * p[i]) % mod;
    else {
      int v = --p[i];
      while (v != 1) {
        int u = pri[v], ci = 0;
        while (v % u == 0) v /= u, ci++;
        if (cnt[u] < ci) {
          for (int j = cnt[u] + 1; j <= ci; j++) ans = (1ll * ans * u) % mod;
          cnt[u] = ci;
          mx[u] = 1;
        } else
          mx[u] += (cnt[u] == ci);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int v = p[i], flag = 0;
    while (v != 1) {
      int u = pri[v], ci = 0;
      while (v % u == 0) v /= u, ci++;
      if (cnt[u] == ci && mx[u] == 1) flag = 1;
    }
    if (!flag) {
      ans++;
      break;
    }
  }
  printf("%d\n", ans);
  return 0;
}