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;
using lint = long long;
const int MAXN = 2000005;
const lint mod = 1e9 + 7;
int n, a[MAXN];
int che[MAXN];
int cnt[MAXN];
int yielder[MAXN];
lint ipow(lint x, lint p) {
  lint ret = 1, piv = x;
  while (p) {
    if (p & 1) ret = ret * piv % mod;
    piv = piv * piv % mod;
    p >>= 1;
  }
  return ret;
}
vector<int> factor(int v) {
  if (v == 1) {
    return vector<int>();
  }
  vector<int> ans;
  while (v > 1) {
    ans.push_back(che[v]);
    v /= che[v];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i < MAXN; i++) {
    for (int j = i; j < MAXN; j += i) {
      if (!che[j]) che[j] = i;
    }
  }
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  set<int> s;
  int useless = 0;
  queue<int> que;
  for (int i = 0; i < n;) {
    int e = i;
    while (e < n && a[i] == a[e]) e++;
    if (e - i == 1)
      s.insert(a[i]);
    else {
      if (e - i >= 3) useless = 1;
      que.push(a[i] - 1);
      que.push(a[i]);
    }
    i = e;
  }
  vector<int> will_check;
  while (!que.empty()) {
    auto x = que.front();
    will_check.push_back(x);
    que.pop();
    auto v = factor(x);
    for (int i = 0; i < v.size();) {
      int e = i;
      while (e < v.size() && v[i] == v[e]) e++;
      if (cnt[v[i]] < e - i) {
        cnt[v[i]] = e - i;
      }
      if (cnt[v[i]] && s.find(v[i]) != s.end()) {
        s.erase(v[i]);
        que.push(v[i] - 1);
      }
      i = e;
    }
  }
  for (auto &i : s) {
    cnt[i] = 1;
  }
  lint ans = 1;
  for (int i = 2; i < MAXN; i++) {
    ans *= ipow(i, cnt[i]);
    ans %= mod;
  }
  if (!useless) {
    for (int i = 0; i < n;) {
      int e = i;
      while (e < n && a[e] == a[i]) e++;
      if (e - i == 2)
        a[i]--;
      else if (s.find(a[i]) == s.end())
        a[i]--;
      i = e;
    }
    for (int i = 0; i < n; i++) {
      auto x = factor(a[i]);
      for (int j = 0; j < x.size();) {
        int e = j;
        while (e < x.size() && x[e] == x[j]) e++;
        if (cnt[x[j]] == e - j) yielder[x[j]]++;
        j = e;
      }
    }
    for (int i = 0; i < n; i++) {
      auto x = factor(a[i]);
      bool useful = 0;
      for (int j = 0; j < x.size();) {
        int e = j;
        while (e < x.size() && x[e] == x[j]) e++;
        if (cnt[x[j]] == e - j && yielder[x[j]] == 1) useful = 1;
        j = e;
      }
      if (!useful) useless = 1;
    }
  }
  ans += useless;
  ans %= mod;
  cout << ans << endl;
}