#include <bits/stdc++.h>
using namespace std;
int p[200010];
bool vis[200010];
int q[2000010];
int c[2000010];
vector<int> d[2000010];
const int N = 2000000;
void init() {
  for (int i = 2; i <= N; i++) {
    if (d[i].size()) continue;
    for (int j = i; j <= N; j += i) d[j].push_back(i);
  }
  return;
}
void work(int x) {
  for (int i = 0; i < d[x].size(); i++) {
    int y = d[x][i];
    int cnt = 0, t = x;
    while (t % y == 0) {
      t /= y;
      cnt++;
    }
    if (cnt > q[y]) {
      q[y] = cnt;
      c[y] = 1;
    } else if (cnt == q[y])
      c[y]++;
  }
  return;
}
bool check(int x) {
  for (int i = 0; i < d[x].size(); i++) {
    int y = d[x][i];
    int cnt = 0, t = x;
    while (t % y == 0) {
      t /= y;
      cnt++;
    }
    if (cnt == q[y] && c[y] == 1) return false;
  }
  return true;
}
const int Mod = 1000000007;
int main() {
  init();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  for (int i = n; i >= 1; i--) {
    int x = p[i];
    if (!q[x]) {
      q[x] = 1;
      c[x] = 1;
    } else {
      vis[i] = true;
      work(p[i] - 1);
    }
  }
  int ans = 1, add = 0;
  for (int i = 1; i <= n; i++)
    if (vis[i] && check(p[i] - 1)) add = 1;
  for (int i = 2; i <= N; i++) {
    int t = q[i];
    for (int j = 0; j < t; j++) ans = 1ll * ans * i % Mod;
  }
  printf("%d\n", (ans + add) % Mod);
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following 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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int p[200010];
bool vis[200010];
int q[2000010];
int c[2000010];
vector<int> d[2000010];
const int N = 2000000;
void init() {
  for (int i = 2; i <= N; i++) {
    if (d[i].size()) continue;
    for (int j = i; j <= N; j += i) d[j].push_back(i);
  }
  return;
}
void work(int x) {
  for (int i = 0; i < d[x].size(); i++) {
    int y = d[x][i];
    int cnt = 0, t = x;
    while (t % y == 0) {
      t /= y;
      cnt++;
    }
    if (cnt > q[y]) {
      q[y] = cnt;
      c[y] = 1;
    } else if (cnt == q[y])
      c[y]++;
  }
  return;
}
bool check(int x) {
  for (int i = 0; i < d[x].size(); i++) {
    int y = d[x][i];
    int cnt = 0, t = x;
    while (t % y == 0) {
      t /= y;
      cnt++;
    }
    if (cnt == q[y] && c[y] == 1) return false;
  }
  return true;
}
const int Mod = 1000000007;
int main() {
  init();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + n + 1);
  for (int i = n; i >= 1; i--) {
    int x = p[i];
    if (!q[x]) {
      q[x] = 1;
      c[x] = 1;
    } else {
      vis[i] = true;
      work(p[i] - 1);
    }
  }
  int ans = 1, add = 0;
  for (int i = 1; i <= n; i++)
    if (vis[i] && check(p[i] - 1)) add = 1;
  for (int i = 2; i <= N; i++) {
    int t = q[i];
    for (int j = 0; j < t; j++) ans = 1ll * ans * i % Mod;
  }
  printf("%d\n", (ans + add) % Mod);
  return 0;
}

```