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;
using ll = long long;
const ll MOD = int(1e9) + 7;
const int MAXV = 2.2e6;
const int V = 2.1e6;
int fac[MAXV];
int N;
int cnt[MAXV];
int has[MAXV];
int best[MAXV];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i <= V; i++) {
    if (fac[i]) continue;
    for (int j = i; j <= V; j += i) {
      fac[j] = i;
    }
  }
  cin >> N;
  for (int i = 0; i < N; i++) {
    int p;
    cin >> p;
    assert(p <= V);
    assert(fac[p] == p);
    cnt[p]++;
  }
  ll res = 1;
  bool extra = false;
  for (int p = V; p >= 2; p--) {
    if (fac[p] != p) continue;
    if (cnt[p] && !has[p]) {
      cnt[p]--;
      has[p] = 1;
      best[p] = 1;
    }
    if (has[p]) {
      for (int i = 0; i < best[p]; i++) {
        res = res * ll(p) % MOD;
      }
    }
    if (!cnt[p]) continue;
    assert(has[p]);
    if (cnt[p] >= 2) extra = true;
    int v = p - 1;
    while (v > 1) {
      int q = fac[v];
      int e = 0;
      while (v % q == 0) v /= q, e++;
      if (e > best[q]) {
        best[q] = e;
        has[q] = 0;
      }
      if (e == best[q]) {
        has[q]++;
      }
    }
  }
  if (!extra) {
    for (int p = V; p >= 2; p--) {
      if (fac[p] != p) continue;
      if (cnt[p]) {
        assert(has[p]);
        int v = p - 1;
        bool good = true;
        while (v > 1) {
          int q = fac[v];
          int e = 0;
          while (v % q == 0) v /= q, e++;
          if (e == best[q] && has[q] == 1) {
            good = false;
            break;
          }
        }
        if (good) {
          extra = true;
          break;
        }
      }
    }
  }
  if (extra) {
    res++;
    res %= MOD;
  }
  cout << res << '\n';
  return 0;
}