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 inf = 1 << 30;
const int MOD = 1e9 + 7;
const int maxn = 2e5 + 7;
int p[maxn], n;
int vis[2000000 + 7], cnt[2000000 + 7];
bool vispi[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + 1 + n);
  for (int i = n; i; i--) {
    vispi[i] = vis[p[i]];
    if (!vis[p[i]])
      vis[p[i]] = 1;
    else {
      int M = p[i] - 1;
      for (int j = 2; j * j <= M; j++) {
        if (M % j == 0) {
          int ind = 0;
          for (; M % j == 0; M /= j) ind++;
          if (ind > vis[j])
            vis[j] = ind, cnt[j] = 1;
          else if (ind == vis[j])
            cnt[j]++;
        }
      }
      if (M > 1) {
        if (!vis[M])
          vis[M] = 1, cnt[M] = 1;
        else if (vis[M] == 1)
          cnt[M]++;
      }
    }
  }
  bool flag = false;
  for (int i = n; i; i--) {
    if (!vispi[i]) {
      if (vis[p[i]] > 1 || (vis[p[i]] == 1 && cnt[p[i]] > 1)) flag = true;
    } else {
      int M = p[i] - 1;
      flag = true;
      for (int j = 2; j * j <= M; j++) {
        if (M % j == 0) {
          int ind = 0;
          for (; M % j == 0; M /= j) ind++;
          if (ind >= vis[j] && !(ind == vis[j] && cnt[j] > 1)) flag = false;
        }
      }
      if (M > 1) {
        if (vis[M] == 1 && !(cnt[M] > 1)) flag = false;
      }
    }
    if (flag) break;
  }
  long long ans = 1;
  for (int i = 1; i <= p[n]; i++)
    while (vis[i]--) ans = ans * i % MOD;
  ans = (ans + flag) % MOD;
  printf("%lld\n", ans);
  return 0;
}