#include <bits/stdc++.h>
int maxp[2000000 + 1];
int lcm[2000000 + 1];
int lcm_cnt[2000000 + 1];
int p[200000];
int q[200000], qp;
int max(int x, int y) { return x > y ? x : y; }
int cmp(const void *a, const void *b) { return *(int *)b - *(int *)a; }
int main(void) {
  for (int i = 0; i <= 2000000; i++) {
    maxp[i] = i;
  }
  for (int i = 2; i <= 2000000; i++) {
    if (maxp[i] == i) {
      for (int j = i * 2; j <= 2000000; j += i) {
        maxp[j] = i;
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
  }
  qsort(p, n, sizeof(int), cmp);
  int one = 0;
  for (int i = 0; i < n; i++) {
    int x = p[i];
    if (lcm[x] == 0) {
      lcm[x] = 1;
      continue;
    }
    q[qp++] = x;
    x--;
    while (x != 1) {
      int p = maxp[x];
      int e = 0;
      while (x % p == 0) {
        x /= p;
        e++;
      }
      if (lcm[p] < e) {
        lcm[p] = e;
        lcm_cnt[p] = 1;
      } else if (lcm[p] == e) {
        lcm_cnt[p]++;
      }
    }
  }
  for (int i = 0; i < qp; i++) {
    int x = q[i] - 1;
    bool forall = true;
    while (x != 1) {
      int p = maxp[x];
      int e = 0;
      while (x % p == 0) {
        x /= p;
        e++;
      }
      if (lcm[p] == e && lcm_cnt[p] == 1) {
        forall = false;
      }
    }
    if (forall) {
      one = 1;
    }
  }
  long long ans = 1;
  for (int i = 0; i <= 2000000; i++) {
    for (int j = 0; j < lcm[i]; j++) {
      ans *= i;
      ans %= 1000000007;
    }
  }
  ans += one;
  ans %= 1000000007;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
int maxp[2000000 + 1];
int lcm[2000000 + 1];
int lcm_cnt[2000000 + 1];
int p[200000];
int q[200000], qp;
int max(int x, int y) { return x > y ? x : y; }
int cmp(const void *a, const void *b) { return *(int *)b - *(int *)a; }
int main(void) {
  for (int i = 0; i <= 2000000; i++) {
    maxp[i] = i;
  }
  for (int i = 2; i <= 2000000; i++) {
    if (maxp[i] == i) {
      for (int j = i * 2; j <= 2000000; j += i) {
        maxp[j] = i;
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
  }
  qsort(p, n, sizeof(int), cmp);
  int one = 0;
  for (int i = 0; i < n; i++) {
    int x = p[i];
    if (lcm[x] == 0) {
      lcm[x] = 1;
      continue;
    }
    q[qp++] = x;
    x--;
    while (x != 1) {
      int p = maxp[x];
      int e = 0;
      while (x % p == 0) {
        x /= p;
        e++;
      }
      if (lcm[p] < e) {
        lcm[p] = e;
        lcm_cnt[p] = 1;
      } else if (lcm[p] == e) {
        lcm_cnt[p]++;
      }
    }
  }
  for (int i = 0; i < qp; i++) {
    int x = q[i] - 1;
    bool forall = true;
    while (x != 1) {
      int p = maxp[x];
      int e = 0;
      while (x % p == 0) {
        x /= p;
        e++;
      }
      if (lcm[p] == e && lcm_cnt[p] == 1) {
        forall = false;
      }
    }
    if (forall) {
      one = 1;
    }
  }
  long long ans = 1;
  for (int i = 0; i <= 2000000; i++) {
    for (int j = 0; j < lcm[i]; j++) {
      ans *= i;
      ans %= 1000000007;
    }
  }
  ans += one;
  ans %= 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```