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 = 2e5 + 5;
const int M = 2e6 + 5;
const int mod = 1e9 + 7;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int mul(int x, int y) { return (long long)x * y % mod; }
int n;
int a[N];
int lp[M];
vector<int> pr;
int cnt[M], many[M];
int main() {
  for (int i = 2; i < M; i++) {
    if (!lp[i]) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (auto x : pr) {
      if (x > lp[i] or x * i >= M) break;
      lp[x * i] = x;
    }
  }
  scanf("%d", &n);
  bool add_one = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    bool flag = 0;
    while (x > 1) {
      int y = lp[x], c = 0;
      while (x % y == 0) {
        x /= y;
        c++;
      }
      if (c > cnt[y]) {
        cnt[y] = c;
        flag = 1;
        many[y] = 1;
      } else if (c == cnt[y])
        many[y]++;
    }
    if (!flag) {
      x = a[i];
      int y = lp[x], c = 0;
      while (x % y == 0) {
        x /= y;
        c++;
      }
      if (c == cnt[y]) many[y]--;
      x = --a[i];
      while (x > 1) {
        int y = lp[x], c = 0;
        while (x % y == 0) {
          x /= y;
          c++;
        }
        if (c > cnt[y]) {
          cnt[y] = c;
          flag = 1;
          many[y] = 1;
        } else if (c == cnt[y])
          many[y]++;
      }
      if (!flag) add_one = 1;
    }
  }
  int ans = 1;
  for (int i = 2; i < M; i++) {
    for (int it = 0; it < cnt[i]; it++) {
      ans = mul(ans, i);
    }
  }
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    bool flag = 0;
    while (x > 1) {
      int y = lp[x], c = 0;
      while (x % y == 0) {
        x /= y;
        c++;
      }
      if (c == cnt[y] and many[y] == 1) {
        flag = 1;
      }
    }
    if (!flag) add_one = 1;
  }
  ans = add(ans, add_one);
  printf("%d\n", ans);
  return 0;
}