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;
long long gcd(long long a, long long b) {
  if (a == 0LL) return b;
  if (b == 0LL) return a;
  return gcd(b, a % b);
}
const int N = 200005;
const int M = 2000006;
int smpf[M];
int cnt[M];
int how[M];
int P[N];
bool mark[N];
void pre() {
  for (int i = 1; i < M; i++) smpf[i] = i;
  for (int i = 2; i < M; i++)
    if (smpf[i] == i)
      for (int j = i + i; j < M; j += i)
        if (smpf[j] == j) smpf[j] = i;
  return;
}
bool benefit(int x) {
  if (x == 1) return false;
  vector<pair<int, int> > v;
  v.clear();
  int cnt1, num;
  while (x != 1) {
    num = smpf[x];
    cnt1 = 0;
    while (smpf[x] == num) {
      cnt1++;
      x = x / num;
    }
    v.push_back(make_pair(num, cnt1));
  }
  bool flag = false;
  for (int i = 0; i < v.size(); i++)
    flag = flag | (v[i].second > cnt[v[i].first]);
  if (flag) {
    for (int i = 0; i < v.size(); i++) {
      if (v[i].second > cnt[v[i].first]) {
        how[v[i].first] = 1;
        cnt[v[i].first] = v[i].second;
      } else if (v[i].second == cnt[v[i].first])
        how[v[i].first]++;
    }
  }
  return flag;
}
bool check(int x) {
  if (x == 1) return false;
  vector<pair<int, int> > v;
  v.clear();
  int cnt1, num;
  while (x != 1) {
    num = smpf[x];
    cnt1 = 0;
    while (smpf[x] == num) {
      cnt1++;
      x = x / num;
    }
    v.push_back(make_pair(num, cnt1));
  }
  bool flag = true;
  for (int i = 0; i < v.size(); i++) {
    if (cnt[v[i].first] == v[i].second && how[v[i].first] == 1) {
      flag = false;
      break;
    }
  }
  return flag;
}
void solve() {
  pre();
  int n;
  bool flag = false;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> P[i];
  sort(P + 1, P + 1 + n, greater<int>());
  for (int i = 1; i <= n; i++) {
    if (benefit(P[i]))
      continue;
    else if (benefit(P[i] - 1)) {
      mark[i] = true;
      continue;
    } else
      flag = true;
  }
  if (!flag) {
    for (int i = 1; i <= n; i++)
      if (mark[i] && check(P[i] - 1)) {
        flag = true;
        break;
      }
  }
  long long ans = 1LL;
  for (int i = 1; i < M; i++) {
    while (cnt[i] > 0) {
      ans = (ans * ((long long)i)) % 1000000007LL;
      cnt[i]--;
    }
  }
  if (flag) ans++;
  ans = ans % 1000000007LL;
  cout << ans << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  clock_t clk = clock();
  int t = 1;
  for (int tests = 1; tests <= t; tests++) {
    solve();
  }
  clk = clock() - clk;
  cerr << "Time Elapsed: " << fixed << setprecision(10)
       << ((long double)clk) / CLOCKS_PER_SEC << "\n";
  return 0;
}