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 = 2000005;
const long long MOD = 1000000007;
int P[200005];
int pfac[MAXN], has[MAXN], tot[MAXN];
vector<pair<int, int>> contrib[200005];
int main() {
  for (int i = 2; i < MAXN; ++i) pfac[i] = i;
  for (int i = 2; i < MAXN; ++i)
    if (pfac[i] == i) {
      for (int j = i + i; j < MAXN; j += i) pfac[j] = i;
    }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &P[i]);
  sort(P, P + n);
  long long ans = 1;
  for (int i = n - 1; i >= 0; --i) {
    if (has[P[i]]) {
      int x = P[i] - 1;
      while (x > 1) {
        int p = pfac[x];
        int cnt = 0;
        while (x % p == 0) {
          x /= p;
          ++cnt;
        }
        if (has[p] < cnt) tot[p] = 0;
        contrib[i].push_back({p, cnt});
        while (has[p] < cnt) {
          ++has[p];
          ans = ans * p % MOD;
        }
        if (has[p] == cnt) ++tot[p];
      }
    } else {
      tot[P[i]] = 1;
      contrib[i].push_back({P[i], 1});
      has[P[i]] = 1;
      ans = ans * P[i] % MOD;
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    bool bad = false;
    for (auto x : contrib[i]) {
      if (x.second == has[x.first]) {
        if (tot[x.first] == 1) bad = true;
      }
    }
    if (!bad) {
      ++ans;
      break;
    }
  }
  ans %= MOD;
  printf("%lld\n", ans);
}