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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = (int)1e9 + 7;
int n;
const int maxN = 2 * (int)1e6 + 100;
int lp[maxN];
int cnt_hs[maxN];
int coef[maxN];
int ans = 1;
int mult(int a, int b) { return (1LL * a * b) % mod; }
vector<int> gg;
bool add(int x) {
  bool f = false;
  int cp = x;
  while (x > 1) {
    int p = lp[x];
    int ex = 0;
    while (x % p == 0) {
      x /= p;
      ex++;
    }
    if (ex > coef[p]) {
      f = true;
      for (int j = coef[p]; j < ex; j++) ans = mult(ans, p);
      coef[p] = ex;
    }
  }
  if (f) gg.push_back(cp);
  return f;
}
int cnt_val[maxN];
void add2(int x) {
  while (x > 1) {
    int p = lp[x];
    int ex = 0;
    while (x % p == 0) {
      x /= p;
      ex++;
    }
    assert(ex <= coef[p]);
    if (ex == coef[p]) cnt_val[p]++;
  }
}
bool can_del(int x) {
  bool f = true;
  while (x > 1) {
    int p = lp[x];
    int ex = 0;
    while (x % p == 0) {
      x /= p;
      ex++;
    }
    if (ex == coef[p] && cnt_val[p] == 1) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 2; i < maxN; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      for (int j = 2 * i; j < maxN; j += i) lp[j] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    cnt_hs[x]++;
  }
  bool go = false;
  for (int i = maxN - 1; i >= 2; i--) {
    if (lp[i] != i) continue;
    if (cnt_hs[i] == 0) continue;
    if (add(i)) {
      cnt_hs[i]--;
    }
    if (cnt_hs[i] > 0) {
      if (add(i - 1)) {
        cnt_hs[i]--;
      }
    }
    if (cnt_hs[i] > 0) go = true;
  }
  if (go) {
    ans = (ans + 1) % mod;
    cout << ans;
  } else {
    for (int v : gg) {
      add2(v);
    }
    for (int v : gg) {
      if (can_del(v)) {
        ans = (ans + 1) % mod;
        cout << ans;
        return 0;
      }
    }
    cout << ans;
  }
  return 0;
}