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;
namespace thecold {
inline int read() {
  int res = 0;
  bool flag = false;
  char c = getchar();
  while (!isdigit(c)) flag = (c == '-'), c = getchar();
  while (isdigit(c)) res = res * 10 + (c ^ 48), c = getchar();
  return !flag ? res : -res;
}
const int Max_n = 2e5 + 5;
const int jzp = 1e9 + 7;
const int Max_d = 2e6 + 5;
int n, a[Max_n], mindiv[Max_d], prime[Max_d], tot;
int m, mx[Max_d], ans = 1, apr[Max_d], cnt[Max_d];
inline void prepare() {
  const int n = 2e6;
  for (int i = 2; i <= n; ++i) {
    if (!mindiv[i]) mindiv[i] = i, prime[++tot] = i;
    for (int j = 1; j <= tot && prime[j] * i <= n; ++j) {
      mindiv[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
inline int quickpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % jzp;
    x = x * x % jzp, y >>= 1;
  }
  return ret;
}
inline int check(const int x, const int flag) {
  if (!flag) return cnt[x] != 1;
  int tmp = x, ret = 1;
  while (tmp != 1) {
    int md = mindiv[tmp], cnt = 0;
    while (tmp % md == 0) tmp /= md, ++cnt;
    if (mx[md] == cnt && apr[md] == 1) ret = 0;
  }
  return ret;
}
inline void main() {
  n = read(), prepare();
  for (int i = 1; i <= n; ++i) a[i] = read(), ++cnt[a[i]];
  sort(a + 1, a + n + 1), reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    bool hav = false;
    int tmp = a[i];
    while (tmp != 1) {
      int md = mindiv[tmp], cnt = 0;
      while (tmp % md == 0) tmp /= md, ++cnt;
      if (mx[md] < cnt) hav = true, mx[md] = cnt, apr[md] = 1;
    }
    if (!hav) {
      tmp = a[i] - 1;
      if (tmp == 1) m = 1;
      while (tmp != 1) {
        int md = mindiv[tmp], cnt = 0;
        while (tmp % md == 0) tmp /= md, ++cnt;
        if (cnt == mx[md]) ++apr[md];
        if (mx[md] < cnt) hav = true, mx[md] = cnt, apr[md] = 1;
      }
    }
  }
  for (int i = 1; i <= 2e6; ++i) ans = 1ll * ans * quickpow(i, mx[i]) % jzp;
  for (int i = 1; i <= n; ++i) m |= (check(a[i], 0) && check(a[i] - 1, 1));
  printf("%d\n", (ans + m) % jzp);
}
}  // namespace thecold
int main() {
  thecold::main();
  return 0;
}