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;
int get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int mo = 1e9 + 7;
const int N = 2e6 + 5;
int cnt[N];
int val[N];
int lst[N];
int pri[N], k;
int n;
int tim[N];
bool bz[N];
int main() {
  n = get();
  for (int i = 1; i <= n; i++) {
    int x = get();
    cnt[x]++;
  }
  lst[1] = 1;
  for (int i = 2; i <= 2e6; i++) {
    if (!lst[i]) pri[++k] = lst[i] = i;
    for (int j = 1; j <= k; j++) {
      if (1ll * pri[j] * i > 2e6) break;
      lst[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
  for (int i = 1; i <= 2e6; i++) val[i] = 1;
  int C = 0;
  for (int i = 2e6; i >= 2; i--) {
    for (int T = 1; T <= cnt[i]; T++) {
      if (val[i] < i)
        bz[i] = 1, tim[i] = 1, val[i] = i;
      else {
        bz[i - 1] = 1;
        int x = i - 1;
        while (x > 1) {
          int v = lst[x];
          int t = 1;
          while (x % v == 0) {
            x /= v;
            t = t * v;
          }
          if (t > val[v]) {
            val[v] = t;
            tim[v] = 1;
          } else if (val[v] == t)
            tim[v]++;
        }
      }
    }
  }
  for (int i = 1; i <= 2e6; i++)
    if (bz[i]) {
      bool can = 1;
      for (int x = i; x > 1;) {
        int v = lst[x], t = 1;
        for (; x % v == 0;) x /= v, t *= v;
        if (val[v] == t && tim[v] == 1) can = 0;
      }
      if (can) {
        C = 1;
        break;
      }
    }
  int ans = 1;
  for (int i = 1; i <= 2e6; i++) ans = 1ll * ans * val[i] % mo;
  ans = (ans + C) % mo;
  cout << ans << endl;
  return 0;
}