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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, a[2000005], f[2000005], g[2000005], b[2000005], p[2000005], c[2000005];
long long ans = 1;
bool ok;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int o;
    scanf("%d", &o);
    ++a[o];
  }
  for (int i = 2000000; i; --i) {
    if (!a[i]) continue;
    if (a[i] > 2) ok = 1;
    if (a[i] > 1) {
      f[i] = f[i - 1] = 1;
      continue;
    }
    bool fl = 0;
    for (int j = i; j <= 2000000; j += i)
      if (f[j]) fl = 1, j = 2000000;
    if (fl)
      f[i - 1] = 1;
    else
      f[i] = 1;
  }
  if (f[1]) ok = 1;
  for (int i = 2; i <= 2000000; ++i) {
    if (p[i]) continue;
    for (int j = i + i; j <= 2000000; j += i) p[j] = 1;
  }
  for (int i = 2000000; i >= 2; --i) {
    if (p[i]) continue;
    p[i] = i;
    int t = i;
    while (t <= 2000000 / i) t *= i, p[t] = i;
  }
  for (int i = 2000000; i > 1; --i) {
    if (p[i] == 1) continue;
    int fl = 0;
    for (int j = i + i; j <= 2000000; j += i)
      if (b[j]) fl = 1, j = 2000000;
    if (fl) continue;
    for (int j = i; j <= 2000000; j += i)
      if (f[j]) ++fl;
    if (fl == 1) {
      for (int j = i; j <= 2000000; j += i)
        if (f[j]) g[j] = 1;
    }
    if (fl) b[i] = 1;
  }
  for (int i = 2; i <= 2000000; ++i)
    if (f[i] && !g[i]) ok = 1;
  for (int i = 2; i <= 2000000; ++i)
    if (b[i]) ans = ans * 1ll * i % 1000000007ll;
  if (ok) ans = (ans + 1ll) % 1000000007ll;
  ans = (ans % 1000000007ll + 1000000007ll) % 1000000007ll;
  printf("%I64d\n", ans);
  return 0;
}