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 maxn = 2e5 + 10, maxm = 2e6 + 10, mod = 1e9 + 7;
int n, p[maxn], m, pme[maxm], p_n, rem[maxm], cnt[maxm], foo[maxm];
pair<int, int> a[maxn];
bool isp[maxm];
vector<int> all;
void pre() {
  int N = maxm - 10;
  for (int i = 2; i <= N; ++i) isp[i] = 1;
  for (int i = 2; i <= N; ++i) {
    if (isp[i]) pme[p_n++] = i, rem[i] = i;
    for (int j = 0; j < p_n; ++j) {
      int p = pme[j], x = p * i;
      if (x > N) break;
      isp[x] = 0;
      rem[x] = p;
      if (i % p == 0) break;
    }
  }
  return;
}
void M(int x) {
  all.push_back(x);
  while (x > 1) {
    int y = rem[x], z = 0;
    while (x % y == 0) ++z, x /= y;
    if (z > cnt[y]) {
      foo[y] = 0;
      cnt[y] = z;
    }
    if (cnt[y] == z) ++foo[y];
  }
  return;
}
int F(int x) {
  while (x > 1) {
    int y = rem[x], z = 0;
    while (x % y == 0) ++z, x /= y;
    if (z == cnt[y] && foo[y] == 1) return 0;
  }
  return 1;
}
int main() {
  pre();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", p + i);
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j + 1 <= n && p[j + 1] == p[i]) ++j;
    a[m++] = make_pair(p[i], j - i + 1);
    i = j;
  }
  sort(a, a + m, [&](pair<int, int> u, pair<int, int> v) {
    return u.second == v.second ? u.first > v.first : u.second > v.second;
  });
  bool fuck = 0;
  for (int i = 0; i < m; ++i) {
    int x = a[i].first, y = a[i].second;
    if (cnt[x]) --x;
    M(x);
    --y;
    if (y) {
      if (cnt[x]) --x;
      M(x);
      --y;
      if (y) {
        fuck = 1;
      }
    }
  }
  int lcm = 1;
  for (int x = 2; x < maxm; ++x) {
    for (int j = 0; j < cnt[x]; ++j) lcm = (long long)lcm * x % mod;
  }
  if (fuck) {
    printf("%d\n", (lcm + 1) % mod);
    return 0;
  }
  for (int x : all) {
    if (F(x)) {
      printf("%d\n", (lcm + 1) % mod);
      return 0;
    }
  }
  printf("%d\n", lcm);
  return 0;
}