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 long long INF = 1e9 + 7;
const int N = 2e6 + 10;
int a[N];
int d[N];
int prime[N];
int key[N];
int cnt[N];
int cnt2[N];
int k = 0;
inline void init() {
  for (int i = 2; i < N; i++) {
    if (!a[i]) prime[++k] = i, d[i] = i;
    for (int j = 1; j <= k && (long long)i * prime[j] < N; j++) {
      d[i * prime[j]] = prime[j];
      a[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 2; i < N; i++) {
    if (i / d[i] % d[i] == 0)
      key[i] = d[i] * key[i / d[i]];
    else
      key[i] = d[i];
  }
}
bool check(int x, int o = 0) {
  int tot = 0;
  int cnt = 0;
  while (x > 1) {
    tot++;
    cnt += cnt2[key[x]] > 0;
    cnt2[key[x]] += o;
    x /= key[x];
  }
  return tot = cnt;
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  int ok = 0;
  for (int i = k; i > 0; i--) {
    int p = prime[i];
    if (cnt[p]) {
      if (!check(p, 0)) {
        check(p, 1), cnt[p]--;
      }
    }
    if (cnt[p] > 1)
      ok = 1;
    else if (p == 2 && cnt[p] > 0)
      ok = 1;
    if (cnt[p]) check(p - 1, 1);
  }
  for (int i = N - 1; i > 1; i--) {
    if (cnt2[i] <= 0) {
      continue;
    }
    int x = i;
    int sum = -cnt2[i];
    while (x > 1) {
      x /= d[x];
      cnt2[x] = -2;
    }
  }
  for (int i = 1; i <= k && !ok; i++) {
    int p = prime[i];
    if (cnt[p] == 0) continue;
    int cnt = 0;
    int tot = 0;
    int x = p - 1;
    while (x > 1) {
      tot++;
      if (cnt2[key[x]] != 1) cnt++;
      x /= key[x];
    }
    if (cnt == tot) ok = 1;
  }
  long long ans = 1;
  for (int i = N - 1; i > 1; i--) {
    if (cnt2[i] <= 0) continue;
    ans = ans * i % INF;
  }
  if (ok) ans++;
  if (ans >= INF) ans -= INF;
  printf("%lld\n", ans);
  return 0;
}