#include <bits/stdc++.h>
const int N = 2000007, P = 1000000007;
int cnt[N], low[N], mx[N], num[N], vis[N];
int mul(int a, int b) { return 1ll * a * b % P; }
int pow(int a, int k) {
  int r = 1;
  for (; k; k >>= 1, a = mul(a, a))
    if (k & 1) r = mul(a, r);
  return r;
}
int read() {
  int x = 0, c = getchar();
  while (isspace(c)) c = getchar();
  while (isdigit(c)) (x *= 10) += c & 15, c = getchar();
  return x;
}
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;) {
    for (p = low[x], c = 0; !(x % p); x /= p, ++c)
      ;
    if (c == mx[p] && num[p] == 1) return 0;
  }
  return 1;
}
int main() {
  int n = read(), m = 0, s = 1, t = 0;
  for (int i = 1, p; i <= n; ++i) m = std::max(m, p = read()), ++cnt[p];
  for (int i = 2; i <= m; ++i)
    if (!low[i])
      for (int j = i; j <= m; j += i)
        if (!low[j]) low[j] = i;
  for (int i = m; i; --i)
    for (int t = cnt[i], x, p, c; t; --t)
      if (mx[i])
        for (vis[i] = 1, x = i - 1; x ^ 1;
             s = mul(s, pow(p, std::max(0, c - mx[p]))), update(p, c))
          for (p = low[x], c = 0; !(x % p); x /= p, ++c)
            ;
      else
        update(i, 1), s = mul(s, i);
  for (int i = m, f; i; --i)
    if (vis[i] && check(i - 1)) t = 1;
  printf("%d", (s + t) % P);
}

### Prompt

Please formulate 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>
const int N = 2000007, P = 1000000007;
int cnt[N], low[N], mx[N], num[N], vis[N];
int mul(int a, int b) { return 1ll * a * b % P; }
int pow(int a, int k) {
  int r = 1;
  for (; k; k >>= 1, a = mul(a, a))
    if (k & 1) r = mul(a, r);
  return r;
}
int read() {
  int x = 0, c = getchar();
  while (isspace(c)) c = getchar();
  while (isdigit(c)) (x *= 10) += c & 15, c = getchar();
  return x;
}
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;) {
    for (p = low[x], c = 0; !(x % p); x /= p, ++c)
      ;
    if (c == mx[p] && num[p] == 1) return 0;
  }
  return 1;
}
int main() {
  int n = read(), m = 0, s = 1, t = 0;
  for (int i = 1, p; i <= n; ++i) m = std::max(m, p = read()), ++cnt[p];
  for (int i = 2; i <= m; ++i)
    if (!low[i])
      for (int j = i; j <= m; j += i)
        if (!low[j]) low[j] = i;
  for (int i = m; i; --i)
    for (int t = cnt[i], x, p, c; t; --t)
      if (mx[i])
        for (vis[i] = 1, x = i - 1; x ^ 1;
             s = mul(s, pow(p, std::max(0, c - mx[p]))), update(p, c))
          for (p = low[x], c = 0; !(x % p); x /= p, ++c)
            ;
      else
        update(i, 1), s = mul(s, i);
  for (int i = m, f; i; --i)
    if (vis[i] && check(i - 1)) t = 1;
  printf("%d", (s + t) % P);
}

```