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;
int m = 2000000, ans, flag;
bool mark[2000007];
int cnt[2000007], vis[2000007];
void work() {
  ans = 1;
  for (int p = 2; p <= m; p++) {
    if (!mark[p]) {
      int mx = 1, flagg = 0;
      for (int j = p; j <= m; j += p) {
        mark[j] = 1;
        if (vis[j]) {
          int t = vis[j], c = 1;
          while (t % p == 0) t /= p, c *= p;
          if (mx == c)
            flagg = 1;
          else if (mx < c) {
            mx = c, flagg = 0;
          }
        }
      }
      for (int j = p; j <= m; j += p) {
        if (vis[j] && (flagg || vis[j] % mx != 0)) {
          while (vis[j] % p == 0) vis[j] /= p;
        }
      }
      ans = (1LL * ans * mx) % 1000000007;
    }
  }
  for (int i = 1; i <= m; i++)
    if (vis[i] == 1) {
      flag = 1;
      return;
    }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ++cnt[x];
  }
  for (int i = m; i >= 2; i--) {
    if (cnt[i] == 0) continue;
    bool mark = 1;
    for (int j = i + i; j <= m; j += i) {
      if (vis[j]) {
        mark = 0;
        break;
      }
    }
    if (mark) {
      vis[i] = i;
      --cnt[i];
    }
    if (cnt[i] > 0) {
      vis[i - 1] = i - 1;
      --cnt[i];
    }
    if (cnt[i] > 0) flag = 1;
  }
  work();
  ans = (ans + flag) % 1000000007;
  printf("%d\n", ans);
  return 0;
}