#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e6 + 5;
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
inline int fpow(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = (long long)res * x % mod;
    k >>= 1;
    x = (long long)x * x % mod;
  }
  return res;
}
int n, s, t;
int a[maxn], low[maxn], mx[maxn], num[maxn], vis[maxn];
void update(int p, int x) {
  if (x > mx[p])
    mx[p] = x, num[p] = 1;
  else
    num[p] += mx[p] == x;
}
int check(int x) {
  for (int p, c; x > 1;) {
    p = low[x];
    c = 0;
    while (x % p == 0) x /= p, ++c;
    if (c == mx[p] && num[p] == 1) return 0;
  }
  return 1;
}
int main() {
  n = gi();
  s = 1;
  ++n;
  --n;
  for (int i = 1; i <= n; ++i) a[i] = gi();
  sort(a + 1, a + n + 1);
  for (int i = 2; i <= a[n]; ++i)
    if (!low[i])
      for (int j = i; j <= a[n]; j += i)
        if (!low[j]) low[j] = i;
  for (int i = n, v, x, p, c; i; --i)
    if (mx[v = a[i]])
      for (vis[i] = 1, x = v - 1; x > 1;
           s = (long long)s * fpow(p, max(0, c - mx[p])) % mod, update(p, c)) {
        p = low[x];
        c = 0;
        while (x % p == 0) x /= p, ++c;
      }
    else
      update(v, 1), s = (long long)s * v % mod;
  for (int i = 1; i <= n; ++i)
    if ((vis[i] && check(a[i] - 1)) || (!vis[i] && check(a[i]))) t = 1;
  printf("%d\n", (s + t) % mod);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 mod = 1e9 + 7;
const int maxn = 2e6 + 5;
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
inline int fpow(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = (long long)res * x % mod;
    k >>= 1;
    x = (long long)x * x % mod;
  }
  return res;
}
int n, s, t;
int a[maxn], low[maxn], mx[maxn], num[maxn], vis[maxn];
void update(int p, int x) {
  if (x > mx[p])
    mx[p] = x, num[p] = 1;
  else
    num[p] += mx[p] == x;
}
int check(int x) {
  for (int p, c; x > 1;) {
    p = low[x];
    c = 0;
    while (x % p == 0) x /= p, ++c;
    if (c == mx[p] && num[p] == 1) return 0;
  }
  return 1;
}
int main() {
  n = gi();
  s = 1;
  ++n;
  --n;
  for (int i = 1; i <= n; ++i) a[i] = gi();
  sort(a + 1, a + n + 1);
  for (int i = 2; i <= a[n]; ++i)
    if (!low[i])
      for (int j = i; j <= a[n]; j += i)
        if (!low[j]) low[j] = i;
  for (int i = n, v, x, p, c; i; --i)
    if (mx[v = a[i]])
      for (vis[i] = 1, x = v - 1; x > 1;
           s = (long long)s * fpow(p, max(0, c - mx[p])) % mod, update(p, c)) {
        p = low[x];
        c = 0;
        while (x % p == 0) x /= p, ++c;
      }
    else
      update(v, 1), s = (long long)s * v % mod;
  for (int i = 1; i <= n; ++i)
    if ((vis[i] && check(a[i] - 1)) || (!vis[i] && check(a[i]))) t = 1;
  printf("%d\n", (s + t) % mod);
  return 0;
}

```