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 sz = 2e6 + 10, inf = 1e9, mod = 1e9 + 7;
int n, m = 0, di[sz], st[sz], be[sz], good = 0, add = 0, i = 0, an = 1;
bool us[sz];
vector<int> dot[sz];
vector<pair<int, int> > cha[sz];
int lg, rg, va;
void ins(int nu, int l, int r) {
  if (l >= lg and r <= rg)
    dot[nu].push_back(va);
  else if (!(l > rg or r < lg)) {
    int mid = (l + r) >> 1;
    ins(nu * 2, l, mid);
    ins(nu * 2 + 1, mid + 1, r);
  }
}
void ins2(int x) {
  if (st[x]) {
    x--;
    while (x > 1) {
      int cu = di[x], q = 0;
      while (x % cu == 0) x /= cu, q++;
      if (q > st[cu]) {
        cha[i].push_back(make_pair(cu, st[cu]));
        st[cu] = q;
        if (st[cu] == be[cu]) good++;
      }
      if (us[cu]) {
        cha[i].push_back({cu, -1}), us[cu] = 0, ins2(cu);
      }
    }
  } else {
    cha[i].push_back({x, 0}), st[x] = 1;
    if (st[x] == be[x]) good++;
    cha[i].push_back({x, -1}), us[x] = 1;
  }
}
void build(int nu, int l, int r) {
  for (int a = 0; a < dot[nu].size(); a++) {
    int x = dot[nu][a];
    ins2(x);
    i++;
  }
  if (l == r) {
    if (good == m) add = 1;
  } else {
    int mid = (l + r) >> 1;
    build(nu * 2, l, mid);
    build(nu * 2 + 1, mid + 1, r);
  }
  for (int a = 0; a < dot[nu].size(); a++) {
    i--;
    while (cha[i].size()) {
      int cu = cha[i].back().first, q = cha[i].back().second;
      if (q == -1)
        us[cu] ^= 1;
      else {
        if (st[cu] == be[cu]) good--;
        st[cu] = q;
      }
      cha[i].pop_back();
    }
  }
}
int main() {
  for (int a = 2; a < sz; a++) di[a] = inf;
  for (int a = 2; a < sz; a++) {
    if (di[a] == inf)
      for (int b = a; b < sz; b += a) di[b] = min(di[b], a);
  }
  int n;
  cin >> n;
  int ar[n];
  for (int a = 0; a < n; a++) scanf("%d", &ar[a]);
  sort(ar, ar + n, greater<int>());
  for (int a = 0; a < n; a++) {
    int x = ar[a];
    lg = 0, rg = a - 1, va = x;
    if (lg <= rg) ins(1, 0, n - 1);
    lg = a + 1, rg = n - 1, va = x;
    if (lg <= rg) ins(1, 0, n - 1);
    if (be[x]) x--;
    while (x > 1) {
      int cu = di[x], q = 0;
      while (x % cu == 0) x /= cu, q++;
      be[cu] = max(be[cu], q);
    }
  }
  for (int a = 0; a < sz; a++) {
    if (be[a]) {
      m++;
      for (int b = 0; b < be[a]; b++) an = (1ll * an * a) % mod;
    }
  }
  build(1, 0, n - 1);
  an = (an + add) % mod;
  cout << an;
}