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 __i__, __j__;
class _Debug {
 public:
  template <typename T>
  _Debug& operator,(T val) {
    cout << val << endl;
    return *this;
  }
};
int p[200000];
int f[2000001];
int num[2000001], yes[2000001], type[200000];
int ways[2000001];
vector<pair<int, int> > factor(int x) {
  vector<pair<int, int> > v;
  while (x > 1) {
    int ff = f[x], c = 0;
    while ((x % ff) == 0) c++, x /= ff;
    v.push_back(make_pair(ff, c));
  }
  return v;
}
queue<int> Q;
int done[2000001];
int main() {
  int i;
  int n;
  scanf("%d", &n);
  memset(yes, -1, sizeof(yes));
  for (i = 0; i < n; i++) scanf("%d", &p[i]);
  sort(p, p + n);
  for (i = 0; i < n; i++) yes[p[i]] = i;
  int j;
  vector<int> pp;
  for (i = 2; i <= 2000000; i++) {
    if (f[i] == 0) {
      for (j = i; j <= 2000000; j += i) f[j] = i;
      pp.push_back(i);
    }
  }
  for (i = 1; i < n; i++) {
    if (p[i - 1] == p[i]) {
      num[p[i]] = max(num[p[i]], 1);
      vector<pair<int, int> > v = factor(p[i] - 1);
      for (j = 0; j < v.size(); j++)
        num[v[j].first] = max(num[v[j].first], v[j].second), Q.push(v[j].first);
      type[i - 1] = 1, type[i] = 2, yes[p[i]] = -1, done[p[i]] = 1;
    }
  }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    if (yes[u] == -1) done[u] = 1;
    if (done[u]) continue;
    done[u] = 1;
    vector<pair<int, int> > v = factor(u - 1);
    for (i = 0; i < v.size(); i++)
      num[v[i].first] = max(num[v[i].first], v[i].second), Q.push(v[i].first);
    type[yes[u]] = 2;
  }
  for (i = 0; i < n; i++) {
    if (!done[p[i]]) {
      assert(num[p[i]] == 0);
      assert(type[i] == 0);
      num[p[i]] = max(num[p[i]], 1), type[i] = 1;
    }
  }
  for (i = 0; i < n; i++) {
    vector<pair<int, int> > v = factor((type[i] == 2) ? (p[i] - 1) : p[i]);
    for (j = 0; j < v.size(); j++) {
      if (num[v[j].first] == v[j].second) ways[v[j].first]++;
    }
  }
  int t = 0;
  for (i = 0; i < n; i++) {
    vector<pair<int, int> > v = factor((type[i] == 2) ? (p[i] - 1) : p[i]);
    int useful = 0;
    for (j = 0; j < v.size(); j++) {
      if ((num[v[j].first] == v[j].second) && (ways[v[j].first] == 1))
        useful = 1;
    }
    if (!useful) {
      t = 1;
      break;
    }
  }
  long long int ans = 1;
  for (i = 2; i <= 2000000; i++) {
    if (num[i] > 0) {
      for (j = 0; j < num[i]; j++) ans *= i, ans %= 1000000007;
    }
  }
  if (t) ans++, ans %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}