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 = 2e6, mod = 1e9 + 7;
int n, a[222222], mx[2222222], can[2222222];
int cnt[2222222], cur[2222222], f, fl;
vector<int> v;
vector<pair<int, int> > g[2222222];
long long ans;
multiset<int> s[2222222];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 2; i <= maxn; i++) {
    if (mx[i]) continue;
    for (int j = i + i; j <= maxn; j += i) mx[j] = i;
  }
  sort(a + 1, a + n + 1);
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]]) {
      a[i]--;
      v.clear();
      while (mx[a[i]]) {
        int x = mx[a[i]];
        v.push_back(x);
        while (a[i] % x == 0) {
          cur[x]++;
          a[i] /= x;
        }
      }
      if (a[i] > 1) {
        v.push_back(a[i]);
        cur[a[i]]++;
      }
      for (int j = 0; j < v.size(); j++) {
        cnt[v[j]] = max(cnt[v[j]], cur[v[j]]);
        g[i].push_back(make_pair(v[j], cur[v[j]]));
        cur[v[j]] = 0;
      }
    } else {
      cnt[a[i]]++;
      g[i].push_back(make_pair(a[i], 1));
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      s[g[i][j].first].insert(g[i][j].second);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      s[g[i][j].first].erase(s[g[i][j].first].find(g[i][j].second));
    }
    fl = 0;
    for (int j = 0; j < g[i].size(); j++) {
      if (can[g[i][j].first] < g[i][j].second &&
          (s[g[i][j].first].empty() ||
           *s[g[i][j].first].rbegin() < g[i][j].second)) {
        fl = 1;
        break;
      }
    }
    if (!fl) {
      f = 1;
      break;
    }
    for (int j = 0; j < g[i].size(); j++) {
      can[g[i][j].first] = max(can[g[i][j].first], g[i][j].second);
    }
  }
  ans = 1;
  for (int i = 2; i <= maxn; i++) {
    for (int j = 1; j <= cnt[i]; j++) {
      ans = ans * (long long)i % mod;
    }
  }
  ans = (ans + (long long)f) % mod;
  printf("%I64d\n", ans);
  return 0;
}