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;
int p[2000001];
int num[2000001];
int rik[2000001];
bool c[2000001];
const int mod = 1e9 + 7;
int s[2000001];
void pre() {
  for (int i = 2; i <= 2000000; i++) {
    if (!p[i]) {
      p[i] = i;
      for (int j = i * 2; j <= 2000000; j += i) {
        if (!p[j]) p[j] = i;
      }
    }
  }
  return;
}
bool sp(int x) {
  bool sq = 0;
  while (x > 1) {
    int j = p[x];
    int ned = 0;
    while (p[x] == j && x > 1) {
      ned++;
      x /= p[x];
    }
    if (ned > s[j]) {
      sq = 1;
      rik[j] = 1;
    } else if (ned == s[j])
      rik[j]++;
    s[j] = max(s[j], ned);
  }
  return sq;
}
int power(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (1LL * ans * a) % mod;
    a = (1LL * a * a) % mod;
    b >>= 1;
  }
  return ans;
}
bool chk(int x) {
  bool sq = 0;
  while (x > 1) {
    int j = p[x];
    int ned = 0;
    while (p[x] == j && x > 1) {
      ned++;
      x /= p[x];
    }
    if (ned == s[j] && rik[j] == 1) sq = 1;
  }
  return sq;
}
int main() {
  int n;
  scanf("%d", &n);
  pre();
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    num[x]++;
  }
  int f = 0;
  for (int i = 2000000; i >= 1; i--) {
    if (num[i] >= 2) {
      if (sp(i - 1) == 0) {
        f = 1;
      } else if (s[i] >= 1)
        f = 1;
      s[i] = max(s[i], 1);
      if (num[i] >= 3) f = 1;
      c[i] = 1;
    }
  }
  for (int i = 2000000; i >= 1; i--) {
    if (num[i] == 1) {
      if (s[i]) {
        if (sp(i - 1) == 0)
          f = 1;
        else
          c[i] = 1;
      } else
        s[i] = 1;
    }
  }
  for (int i = 2000000; i >= 1; i--) {
    if (c[i]) {
      if (chk(i - 1) == 0) f = 1;
    }
  }
  int ans = 1;
  for (int i = 1; i <= 2000000; i++) {
    if (s[i]) ans = (1LL * ans * power(i, s[i])) % mod;
  }
  cout << (ans + f) % mod << endl;
  return 0;
}