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 MAXN = 2e5 + 5;
const int MAXV = 2e6 + 5;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int n, tot, prime[MAXV], f[MAXV];
int a[MAXN], Max[MAXV], cnt[MAXV];
bool type[MAXN];
void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (f[i] == 0) prime[++tot] = f[i] = i;
    for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
      int tmp = prime[j] * i;
      if (tmp > n) break;
      f[tmp] = prime[j];
    }
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  init(2e6);
  sort(a + 1, a + n + 1);
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (Max[a[i]]) {
      type[i] = true;
      int tmp = a[i] - 1;
      while (tmp != 1) {
        int fac = f[tmp], now = 0;
        while (tmp % fac == 0) {
          tmp /= fac;
          now++;
        }
        if (now > Max[fac]) {
          Max[fac] = now;
          cnt[fac] = 1;
        } else
          cnt[fac] += Max[fac] == now;
      }
    } else
      Max[a[i]] = cnt[a[i]] = 1;
  }
  int ans = 1;
  for (int i = 1; i <= 2e6; i++)
    for (int j = 1; j <= Max[i]; j++) ans = 1ll * ans * i % P;
  for (int i = 1; i <= n; i++) {
    bool found = false;
    if (type[i]) {
      found = true;
      int tmp = a[i] - 1;
      while (tmp != 1) {
        int fac = f[tmp], now = 0;
        while (tmp % fac == 0) {
          tmp /= fac;
          now++;
        }
        found &= Max[fac] > now || cnt[fac] >= 2;
      }
    } else
      found = Max[a[i]] >= 2 || cnt[a[i]] >= 2;
    if (found) {
      writeln((ans + 1) % P);
      return 0;
    }
  }
  writeln(ans);
  return 0;
}