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>
const int U = 2e6, mo = 1e9 + 7;
int p[U / 10], mn[U + 5], pcnt, i, j, x, n, a[200005], c[U + 5], c2[U + 5],
    ans = 1, zz;
bool b[U + 5];
void ins(int x) {
  bool fl = 0;
  for (; x > 1;) {
    int y = mn[x], z = 0;
    for (; x % y == 0; x /= y, ++z)
      ;
    if (z > c[y])
      c[y] = z, c2[y] = 1, fl = 1;
    else if (z == c[y])
      ++c2[y];
  }
  if (!fl) zz = 1;
}
bool ck(int x) {
  bool fl = 0;
  for (; x > 1;) {
    int y = mn[x], z = 0;
    for (; x % y == 0; x /= y, ++z)
      ;
    if (z == c[y] && c2[y] == 1) fl = 1;
  }
  return fl;
}
int main() {
  for (i = 2; i <= U; ++i) {
    if (!b[i]) p[++pcnt] = i, mn[i] = i;
    for (j = 1; j <= pcnt; ++j) {
      x = i * p[j];
      if (x > U) break;
      b[x] = 1;
      mn[x] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  std::sort(a + 1, a + n + 1);
  for (i = n; i; --i) {
    if (c[a[i]]) --a[i];
    ins(a[i]);
  }
  for (i = 1; i <= n; ++i)
    if (!ck(a[i])) zz = 1;
  for (i = 2; i <= U; ++i)
    for (j = 1; j <= c[i]; ++j) ans = 1ll * ans * i % mo;
  printf("%d\n", (ans + zz) % mo);
  return 0;
}