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 N = 2e6 + 2;
const int mod = 1e9 + 7;
int ar[N], uoc[N], cnt[N], pw[N];
bool haiz[N];
int mul(int x, int y) { return (1ll * x * y) % mod; }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, i, j, k, l, ok = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  for (i = 2; i < N; i++) {
    if (!uoc[i]) {
      uoc[i] = i;
      for (j = i; j < N; j += i) {
        if (!uoc[j]) {
          uoc[j] = i;
        }
      }
    }
  }
  sort(ar + 1, ar + 1 + n);
  for (i = n; i > 0; i--) {
    if (!pw[ar[i]]) {
      pw[ar[i]] = cnt[ar[i]] = 1;
      haiz[i] = true;
    } else {
      if (ar[i] == 2) {
        ok = 1;
      } else {
        j = ar[i] - 1;
        while (j > 1) {
          k = uoc[j];
          l = 0;
          while (j % k == 0) {
            j /= k;
            l++;
          }
          if (pw[k] < l) {
            pw[k] = l;
            cnt[k] = 0;
          }
          if (pw[k] == l) {
            cnt[k]++;
          }
        }
      }
    }
  }
  for (i = n; i > 0; i--) {
    if (haiz[i]) {
      continue;
    }
    if (ar[i] == 2) {
      ok = 1;
    } else {
      bool cac = true;
      j = ar[i] - 1;
      while (j > 1) {
        k = uoc[j];
        l = 0;
        while (j % k == 0) {
          j /= k;
          l++;
        }
        if (pw[k] == l && cnt[k] == 1) {
          cac = false;
          break;
        }
      }
      if (cac) {
        ok = 1;
      }
    }
  }
  k = 1;
  for (i = 1; i < N; i++) {
    for (j = 1; j <= pw[i]; j++) {
      k = mul(k, i);
    }
  }
  if (ok) {
    k++;
    k %= mod;
  }
  cout << k;
}