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 MOD = 1e9 + 7;
const int MAX = 2e6;
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> primes;
  vector<int> lp(MAX + 1, 0);
  for (int i = 2; i <= MAX; i++) {
    if (not lp[i]) {
      lp[i] = i;
      primes.push_back(i);
    }
    for (auto p : primes) {
      if (p > lp[i] or i * p > MAX) break;
      lp[i * p] = p;
    }
  }
  vector<int> p(MAX + 1, 0);
  for (int i = 0; i < N; i++) {
    int t;
    cin >> t;
    p[t]++;
  }
  vector<int> q(MAX + 1, 0);
  vector<int> cnt(MAX + 1, 0);
  int TTT = 0;
  for (int i = MAX; i >= 1; i--) {
    if (p[i] >= 2) {
      if (q[i] < 1) {
        q[i] = 1;
        cnt[i] = 1;
      } else if (q[i] == 1)
        cnt[i]++;
      else
        TTT = 1;
      if (i == 2 || p[i] >= 3) TTT = 1;
      int t = i - 1;
      for (int lastp = lp[t], lpcnt = 0; t > 1; t /= lp[t]) {
        if (lastp == lp[t])
          lpcnt++;
        else {
          lastp = lp[t];
          lpcnt = 1;
        }
        if (q[lastp] < lpcnt) {
          q[lastp] = lpcnt;
          cnt[lastp] = 1;
        } else if (q[lastp] == lpcnt)
          cnt[lastp]++;
      }
    } else if (p[i] == 1) {
      if (q[i] == 0)
        q[i] = 1;
      else {
        if (i == 2) TTT = 1;
        int t = i - 1;
        for (int lastp = lp[t], lpcnt = 0; t > 1; t /= lp[t]) {
          if (lastp == lp[t])
            lpcnt++;
          else {
            lastp = lp[t];
            lpcnt = 1;
          }
          if (q[lastp] < lpcnt) {
            q[lastp] = lpcnt;
            cnt[lastp] = 1;
          } else if (q[lastp] == lpcnt)
            cnt[lastp]++;
        }
      }
    }
  }
  for (int i = MAX; i >= 1; i--) {
    if (p[i] >= 2 or (p[i] >= 1 and (q[i] > 1 or cnt[i] > 1))) {
      int t = i - 1;
      int ttt = 1;
      for (int lastp = lp[t], lpcnt = 0; t > 1; t /= lp[t]) {
        if (lastp == lp[t])
          lpcnt++;
        else {
          lastp = lp[t];
          lpcnt = 1;
        }
        if (q[lastp] == lpcnt and cnt[lastp] == 1) ttt = 0;
      }
      TTT |= ttt;
    }
  }
  int ans = 1;
  for (int i = 1; i <= MAX; i++)
    while (q[i]--) ans = ans * 1ll * i % MOD;
  cout << (ans + TTT) % MOD << endl;
  return 0;
}