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 = 200000, maxA = maxN * 10, mod = 1000000007;
int in[maxN], apd[maxA], T[maxA], P[maxA], multipleM[maxA], dcd[maxN];
inline void mult_mod(int& a, int b) { a = 1LL * a * b % mod; }
void updt(int p, int k) {
  if (k < T[p]) return;
  if (k == T[p]) {
    multipleM[p] = true;
    return;
  }
  for (; T[p] < k; T[p]++) mult_mod(P[p], p);
  multipleM[p] = false;
}
int main() {
  for (int(i) = (1); (i) < (maxA); (i)++) apd[i] = i, P[i] = 1;
  for (int i = 2; i * i < maxA; i++) {
    if (apd[i] != i) continue;
    for (int j = i * i; j < maxA; j += i) apd[j] = i;
  }
  int n;
  scanf("%d", &n);
  for (int(i) = (0); (i) < (n); (i)++) scanf("%d", in + i);
  sort(in, in + n);
  for (int i = n - 1; i >= 0; i--) {
    int a = in[i];
    if (T[a] == 0 or a == 2) {
      updt(a, 1);
      continue;
    }
    a--;
    while (a != 1) {
      int p = apd[a], c = 0;
      for (; a % p == 0; a /= p) c++;
      updt(p, c);
    }
    dcd[i] = true;
  }
  int res = 1;
  for (int(i) = (1); (i) < (maxA); (i)++) mult_mod(res, P[i]);
  bool incr = false;
  for (int i = n - 1; i >= 0; i--) {
    int a = in[i];
    if (!dcd[i]) {
      if (T[a] == 1 and !multipleM[a]) continue;
      incr = true;
      break;
    }
    bool ncs = false;
    a--;
    while (a != 1) {
      int p = apd[a], c = 0;
      for (; a % p == 0; a /= p) c++;
      if (T[p] == c and !multipleM[p]) {
        ncs = true;
        break;
      }
    }
    if (!ncs) {
      incr = true;
      break;
    }
  }
  if (incr) res = (res + 1) % mod;
  printf("%d\n", res);
  return 0;
}