#include <bits/stdc++.h>
using namespace std;
const int inf = 0x7fffffff;
const double eps = 1e-10;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch;
  ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch & 15);
    ch = getchar();
  }
  if (f)
    return x;
  else
    return -x;
}
const int N = 2e6 + 10;
const int mod = 1e9 + 7;
int n, a[N], cnt[N], mx[N], ans = 1, prime[N], val[N];
bool used[N];
void Lin_Sie() {
  used[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!used[i]) prime[++prime[0]] = i, val[i] = i;
    for (int j = 1; j <= prime[0] && 1ll * i * prime[j] < N; j++) {
      used[i * prime[j]] = 1;
      val[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  n = read();
  Lin_Sie();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int i = n; i >= 1; i--) {
    if (!cnt[a[i]])
      cnt[a[i]]++, mx[a[i]] = 1, ans = (1ll * ans * a[i]) % mod;
    else {
      int v = --a[i];
      while (v != 1) {
        int u = val[v], ls = 0;
        while (v % u == 0) v /= u, ls++;
        if (cnt[u] < ls) {
          for (int j = cnt[u] + 1; j <= ls; j++) ans = (1ll * ans * u) % mod;
          cnt[u] = ls;
          mx[u] = 1;
        } else if (cnt[u] == ls)
          mx[u]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int v = a[i], flag = 0;
    while (v != 1) {
      int u = val[v], ls = 0;
      while (v % u == 0) v /= u, ls++;
      if (cnt[u] == ls && mx[u] == 1) flag = 1;
    }
    if (!flag) {
      ans++;
      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 inf = 0x7fffffff;
const double eps = 1e-10;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch;
  ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch & 15);
    ch = getchar();
  }
  if (f)
    return x;
  else
    return -x;
}
const int N = 2e6 + 10;
const int mod = 1e9 + 7;
int n, a[N], cnt[N], mx[N], ans = 1, prime[N], val[N];
bool used[N];
void Lin_Sie() {
  used[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!used[i]) prime[++prime[0]] = i, val[i] = i;
    for (int j = 1; j <= prime[0] && 1ll * i * prime[j] < N; j++) {
      used[i * prime[j]] = 1;
      val[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  n = read();
  Lin_Sie();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int i = n; i >= 1; i--) {
    if (!cnt[a[i]])
      cnt[a[i]]++, mx[a[i]] = 1, ans = (1ll * ans * a[i]) % mod;
    else {
      int v = --a[i];
      while (v != 1) {
        int u = val[v], ls = 0;
        while (v % u == 0) v /= u, ls++;
        if (cnt[u] < ls) {
          for (int j = cnt[u] + 1; j <= ls; j++) ans = (1ll * ans * u) % mod;
          cnt[u] = ls;
          mx[u] = 1;
        } else if (cnt[u] == ls)
          mx[u]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int v = a[i], flag = 0;
    while (v != 1) {
      int u = val[v], ls = 0;
      while (v % u == 0) v /= u, ls++;
      if (cnt[u] == ls && mx[u] == 1) flag = 1;
    }
    if (!flag) {
      ans++;
      break;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```