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;
int cnt[2000005];
bool gt[2000005];
bool com[2000005], nd[2000005];
int main() {
  int n, x;
  bool mo1 = false;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &x);
    cnt[x]++;
  }
  for (int i = 2000000; i >= 1; i--) {
    if (cnt[i] == 0) continue;
    if (cnt[i] > 2) mo1 = true;
    bool hv = false;
    for (int j = i; j <= 2000000; j += i)
      if (gt[j]) hv = true;
    if (hv || cnt[i] > 1) gt[i - 1] = true;
    if (!hv || cnt[i] > 1) gt[i] = true;
  }
  long long ans = 1;
  for (int i = 2; i <= 2000000; i++) {
    if (com[i]) continue;
    int tot = 0, tt = 0;
    for (int j = i; j <= 2000000; j += i) {
      if (!gt[j]) continue;
      int cnt = 0, cj = j;
      do {
        cj /= i;
        cnt++;
      } while (cj % i == 0);
      tot = max(tot, cnt);
    }
    for (int j = i; j <= 2000000; j += i) {
      if (!gt[j]) continue;
      int cnt = 0, cj = j;
      do {
        cj /= i;
        cnt++;
      } while (cj % i == 0);
      if (tot == cnt) tt++;
    }
    for (int j = i; j <= 2000000; j += i) {
      if (!gt[j]) continue;
      int cnt = 0, cj = j;
      do {
        cj /= i;
        cnt++;
      } while (cj % i == 0);
      if (tot == cnt && tt == 1) nd[j] = true;
    }
    for (int j = 0; j < (int)(tot); j++) ans = ans * i % MOD;
    for (int j = i + i; j <= 2000000; j += i) com[j] = true;
  }
  for (int i = 1; i <= 2000000; i++)
    if (gt[i] && !nd[i]) mo1 = true;
  printf("%I64d\n", (ans + mo1) % MOD);
  return 0;
}