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 bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool smin(T &a, const T &b) {
  return a > b ? a = b : a;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b : a;
}
const int N = (int)4e6 + 5, mod = (int)1e9 + 7;
int cnt[N], mx[N], pr[N], mark[N];
int use[N];
int main() {
  for (int j = 2; j < N; ++j)
    if (pr[j] == 0)
      for (int i = j; i < N; i += j) pr[i] = j;
  int n;
  cin >> n;
  set<int> st;
  int add = 0;
  for (int j = 0; j < n; ++j) {
    int first;
    cin >> first;
    cnt[first]++;
    mx[first] = 1;
    use[first]++;
    if (cnt[first] > 1) {
      if (cnt[first] > 2) add = 1;
      mark[first] = 1;
      st.insert(-first);
    }
  }
  set<int> all;
  for (int j = 0; j < N; ++j)
    if (cnt[j]) all.insert(j);
  while ((int)st.size()) {
    int cur = -(*(st.begin()));
    st.erase(-cur);
    mark[cur] = 1;
    if (mx[cur] == 1) use[cur]--;
    cur--;
    all.insert(cur);
    int used = 0;
    while (cur != 1) {
      int p = pr[cur], cnt = 0;
      while (cur % p == 0) cur /= p, ++cnt;
      if (cnt >= 1) {
        if (cnt > mx[p]) used = 1;
        if (cnt > mx[p]) use[p] = 0;
        mx[p] = max(mx[p], cnt);
        if (cnt == mx[p]) use[p]++;
        if (::cnt[p] >= 1) {
          if (!mark[p]++) st.insert(-p);
        }
      }
    }
  }
  for (int j : all) {
    int cur = j;
    int used = 0;
    while (cur != 1) {
      int p = pr[cur], cnt = 0;
      while (cur % p == 0) cur /= p, ++cnt;
      if (cnt == mx[p] && use[p] == 1) used = 1;
    }
    if (!used) add = 1;
  }
  for (int j = 2; j < N; ++j)
    if (cnt[j] >= 1) mx[j] = max(mx[j], 1);
  int res = 1;
  for (int j = 1; j < N; ++j)
    while (mx[j]--) res = (long long)res * j % mod;
  cout << (add + res) % mod << endl;
}