#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 7;
int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
int prm[N], pm[N], tt, n, a[N], c[N][2];
bool v[N];
int main() {
  for (int i = 2; i <= N - 5; ++i) {
    if (!pm[i]) pm[i] = prm[++tt] = i;
    for (int j = 1; i * prm[j] <= N - 5; ++j) {
      pm[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
  n = rd();
  for (int i = 1; i <= n; ++i) a[i] = rd();
  sort(a + 1, a + n + 1), reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (!c[a[i]][0])
      v[i] = 1, c[a[i]][0] = c[a[i]][1] = 1;
    else {
      int x = a[i] - 1;
      while (x > 1) {
        int np = pm[x], cn = 0;
        while (pm[x] == np) ++cn, x /= np;
        if (c[np][0] < cn)
          c[np][0] = cn, c[np][1] = 1;
        else
          c[np][1] += c[np][0] == cn;
      }
    }
  }
  int ans = 1;
  for (int i = 2; i <= N - 5; ++i)
    for (int j = 1; j <= c[i][0]; ++j) ans = 1ll * ans * i % mod;
  for (int i = 1; i <= n; ++i)
    if (!v[i]) {
      int x = a[i] - 1, o = 1;
      while (x > 1) {
        int np = pm[x], cn = 0;
        while (pm[x] == np) ++cn, x /= np;
        o &= cn < c[np][0] || c[np][1] > 1;
      }
      if (o) {
        ans = (ans + o) % mod;
        break;
      }
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 2e6 + 10, mod = 1e9 + 7;
int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
int prm[N], pm[N], tt, n, a[N], c[N][2];
bool v[N];
int main() {
  for (int i = 2; i <= N - 5; ++i) {
    if (!pm[i]) pm[i] = prm[++tt] = i;
    for (int j = 1; i * prm[j] <= N - 5; ++j) {
      pm[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
    }
  }
  n = rd();
  for (int i = 1; i <= n; ++i) a[i] = rd();
  sort(a + 1, a + n + 1), reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (!c[a[i]][0])
      v[i] = 1, c[a[i]][0] = c[a[i]][1] = 1;
    else {
      int x = a[i] - 1;
      while (x > 1) {
        int np = pm[x], cn = 0;
        while (pm[x] == np) ++cn, x /= np;
        if (c[np][0] < cn)
          c[np][0] = cn, c[np][1] = 1;
        else
          c[np][1] += c[np][0] == cn;
      }
    }
  }
  int ans = 1;
  for (int i = 2; i <= N - 5; ++i)
    for (int j = 1; j <= c[i][0]; ++j) ans = 1ll * ans * i % mod;
  for (int i = 1; i <= n; ++i)
    if (!v[i]) {
      int x = a[i] - 1, o = 1;
      while (x > 1) {
        int np = pm[x], cn = 0;
        while (pm[x] == np) ++cn, x /= np;
        o &= cn < c[np][0] || c[np][1] > 1;
      }
      if (o) {
        ans = (ans + o) % mod;
        break;
      }
    }
  printf("%d\n", ans);
  return 0;
}

```