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;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 2e5 + 7;
const int M = 2e6 + 7;
int fast(long long a, long long n, int mod) {
  if (n == 0) return 1;
  if (n & 1) return a * fast(a, n - 1, mod) % mod;
  return fast(a * a % mod, n / 2, mod);
}
int n, T[N], fa[M], is[N];
map<int, pair<int, int> > MP;
void update(int p, int cnt) {
  if (MP[p].first < cnt)
    MP[p] = {cnt, 1};
  else if (MP[p].first == cnt)
    MP[p].second++;
}
bool check(int p, int cnt) {
  if (MP[p].first > cnt)
    return true;
  else if (MP[p].first == cnt)
    return MP[p].second > 1;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i * i < M; i++) {
    if (fa[i] != 0) continue;
    for (int j = i * i; j < M; j += i) {
      if (fa[j] == 0) {
        fa[j] = i;
      }
    }
  }
  for (int i = 2; i < M; i++) {
    if (fa[i] == 0) {
      fa[i] = i;
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> T[i];
  sort(T, T + n);
  reverse(T, T + n);
  for (int i = 0; i < n; i++) {
    int p = T[i];
    if (MP[p].first == 0) {
      update(p, 1);
      is[i] = true;
    } else {
      p--;
      while (p > 1) {
        int cnt = 0, d = fa[p];
        while (p % d == 0) {
          p /= d;
          cnt++;
        }
        update(d, cnt);
      }
    }
  }
  bool can = false;
  for (int i = 0; i < n; i++) {
    int p = T[i];
    if (is[i]) {
      if (check(p, 1)) {
        can = true;
      }
    } else {
      bool ok = true;
      p--;
      while (p > 1) {
        int cnt = 0, d = fa[p];
        while (p % d == 0) {
          p /= d;
          cnt++;
        }
        if (!check(d, cnt)) {
          ok = false;
        }
      }
      if (ok) {
        can = true;
      }
    }
  }
  const int MOD = 1e9 + 7;
  long long ans = 1;
  for (auto p : MP) {
    ans = (ans * fast(p.first, p.second.first, MOD)) % MOD;
  }
  if (can) {
    ans = (ans + 1) % MOD;
  }
  cout << ans << "\n";
  return 0;
}