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 MAXP = 2e6 + 5;
const int MOD = 1e9 + 7;
int N;
int p[MAXP];
int cnt[MAXP];
vector<int> adj[MAXP];
vector<vector<pair<int, int> > > all;
int mx1[MAXP], mx2[MAXP];
void load() {
  scanf("%d", &N);
  while (N--) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
}
inline int mul(int x, int y) { return (long long)x * y % MOD; }
void add(int x) {
  vector<pair<int, int> > f;
  while (x > 1) {
    int t = p[x], ex = 0;
    for (; !(x % t); x /= t) ex++;
    if (ex >= mx1[t]) {
      mx2[t] = mx1[t];
      mx1[t] = ex;
    } else if (ex > mx2[t])
      mx2[t] = ex;
    f.push_back({t, ex});
  }
  all.push_back(f);
}
void dfs(int x) {
  cnt[x] = 0;
  add(x - 1);
  for (auto it : adj[x])
    if (cnt[it]) dfs(it);
}
int solve() {
  for (int i = 2; i < MAXP; i++)
    if (!p[i])
      for (int j = i; j < MAXP; j += i)
        if (!p[j]) p[j] = i;
  for (int i = 2; i < MAXP; i++)
    if (cnt[i])
      for (int j = i + 1; j < MAXP; j += i)
        if (cnt[j]) adj[j].push_back(i);
  bool ok = false;
  for (int i = MAXP - 1; i; i--)
    if (cnt[i]) {
      add(i);
      ok |= cnt[i] > 2;
      if (cnt[i] > 1) dfs(i);
    }
  for (auto it : all) {
    bool tmp = true;
    for (auto curr : it) tmp &= mx2[curr.first] >= curr.second;
    ok |= tmp;
  }
  int sol = 1;
  for (int i = 1; i < MAXP; i++)
    while (mx1[i]--) sol = mul(sol, i);
  return (sol + ok) % MOD;
}
int main() {
  load();
  printf("%d\n", solve());
  return 0;
}