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 MX = 2000010;
int exp(int x, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * x % mod;
    x = 1LL * x * x % mod;
    n >>= 1;
  }
  return ret;
}
vector<int> pr;
int pchk[MX];
void sieve() {
  for (int i = 2; i < MX; i++) {
    if (!pchk[i]) {
      pr.push_back(i);
      pchk[i] = i;
    }
    for (int j = 0; j < pr.size() && i * pr[j] < MX; j++) {
      pchk[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) break;
    }
  }
}
int N;
int cnt[MX], flag[MX];
pair<int, int> pcnt[MX];
void update(int p, int v) {
  if (pcnt[p].first == v)
    pcnt[p].second++;
  else if (pcnt[p].first < v)
    pcnt[p] = pair<int, int>(v, 1);
}
void f(int t) {
  int p = -1;
  int c = 0;
  while (1) {
    if (p != pchk[t]) {
      if (p != -1 && pcnt[p].first <= c) update(p, c);
      p = pchk[t];
      c = 0;
    }
    c++;
    if (t == 1) break;
    t /= pchk[t];
  }
}
int g(int t) {
  int ret = 0;
  int p = -1;
  int c = 0;
  while (1) {
    if (p != pchk[t]) {
      if (p != -1 && pcnt[p].first == c && pcnt[p].second == 1) ret = 1;
      p = pchk[t];
      c = 0;
    }
    c++;
    if (t == 1) break;
    t /= pchk[t];
  }
  return ret;
}
int main() {
  sieve();
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    int t;
    scanf("%d", &t);
    cnt[t]++;
  }
  int add = 0;
  for (int i = 0; i < pr.size(); i++) {
    if (cnt[pr[i]] > 2) {
      update(pr[i], 1);
      f(pr[i] - 1);
      add = 1;
    }
  }
  for (int i = (int)pr.size() - 1; i >= 0; i--) {
    if (cnt[pr[i]] == 2) {
      update(pr[i], 1);
      f(pr[i] - 1);
    }
    if (cnt[pr[i]] == 1) {
      if (pcnt[pr[i]].first) {
        f(pr[i] - 1);
        flag[pr[i]] = 1;
      } else
        update(pr[i], 1);
    }
  }
  if (!add) {
    for (int i = (int)pr.size() - 1; i >= 0; i--) {
      if (cnt[pr[i]] == 2) {
        if (!g(pr[i] - 1)) add = 1;
      }
      if (cnt[pr[i]] == 1) {
        if (flag[pr[i]] && !g(pr[i] - 1)) add = 1;
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < pr.size(); i++) {
    ans = 1LL * ans * exp(pr[i], pcnt[pr[i]].first) % mod;
  }
  ans += add;
  ans %= mod;
  printf("%d", ans);
}