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;
const int N = 2e6 + 50;
const long long mod = 1e9 + 7;
inline long long Pow(long long x, long long exp) {
  long long res = 1;
  for (; exp; exp >>= 1, x = x * x % mod)
    if (exp & 1) res = res * x % mod;
  return res;
}
int pri[N / 3], m;
bool notp[N];
vector<int> divi[N];
inline void Init() {
  for (int i = 2; i < N; i++) divi[i].reserve(8);
  for (int i = 2; i < N; i++)
    if (!notp[i]) {
      pri[++m] = i;
      divi[i].push_back(i);
      for (int j = i + i; j < N; j += i) {
        notp[j] = 1;
        divi[j].push_back(i);
      }
    }
}
int n, a[N], cnt[N], num[N], state[N];
inline void Insert(int x) {
  int t = x;
  for (auto i : divi[x]) {
    int c = 0;
    while (t % i == 0) {
      t /= i;
      c++;
    }
    if (cnt[i] < c)
      cnt[i] = c, num[i] = 1;
    else if (cnt[i] == c)
      num[i]++;
  }
}
inline bool check(int x) {
  int t = x;
  for (auto i : divi[x]) {
    int c = 0;
    while (t % i == 0) {
      t /= i;
      c++;
    }
    if (cnt[i] == c && num[i] == 1) return 0;
  }
  return 1;
}
int main() {
  Init();
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    a[x]++;
  }
  bool add = 0;
  for (int i = 1; i < N; i++) {
    if (a[i] > 2) add = 1;
    if (a[i] >= 2) {
      Insert(i);
      Insert(i - 1);
      state[i] = 3;
    }
  }
  for (int i = N - 1; i; i--)
    if (a[i] == 1) {
      if (cnt[i])
        Insert(i - 1), state[i] = 1;
      else
        Insert(i), state[i] = 2;
    }
  for (int i = 1; i < N && !add; i++) {
    if (state[i] & 1) add |= check(i - 1);
    if (state[i] & 2) add |= check(i);
  }
  long long ans = 1;
  for (int i = 2; i < N; i++)
    if (cnt[i]) ans = ans * Pow(i, cnt[i]) % mod;
  printf("%lld\n", (ans + add) % mod);
  return 0;
}