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;
using i64 = long long;
const int maxN = 2234567;
int n, x;
int e[maxN], cnt[maxN], p[maxN], cc[maxN];
bool st[maxN];
void add(int x) {
  int y = x;
  while (y != 1) {
    int p1 = p[y];
    int e1 = 0;
    while (y % p1 == 0) {
      y /= p1;
      e1++;
    }
    if (e[p1] < e1) {
      e[p1] = e1;
      cc[p1] = 0;
    }
    if (e[p1] == e1) cc[p1]++;
  }
}
bool check(int x) {
  int y = x;
  while (y != 1) {
    int p1 = p[y];
    int e1 = 0;
    while (y % p1 == 0) {
      y /= p1;
      e1++;
    }
    if (e[p1] == e1 && cc[p1] == 1) return 0;
  }
  return 1;
}
bool checkMixed() {
  for (int i = 2000000; i >= 2; i--) {
    if (cnt[i] == 0 || st[i] == 0) continue;
    if (cnt[i] >= 3) return 1;
    if (check(i - 1)) return 1;
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    cnt[x]++;
  }
  for (int i = 2; i <= 2000000; i++) {
    if (p[i] == 0) {
      for (int j = i; j <= 2000000; j += i) {
        if (p[j] == 0) {
          p[j] = i;
        }
      }
    }
  }
  for (int i = 2000000; i >= 2; i--) {
    if (cnt[i] == 0) continue;
    if (e[i] == 0) {
      if (cnt[i] >= 2)
        add(i), add(i - 1), st[i] = 1;
      else
        add(i);
    } else {
      add(i - 1), st[i] = 1;
    }
  }
  i64 ans = 1;
  for (int i = 2; i <= 2000000; i++) {
    for (int j = 0; j < e[i]; j++) ans = ans * i % 1000000007;
  }
  if (checkMixed()) ans = (ans + 1) % 1000000007;
  printf("%lld\n", ans);
}