#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005, maxp = 2000005, mod = 1000000007;
int n, ps, lcm, extra;
int p[maxn], minn[maxp], P[maxp], c[maxp], maxx[maxp], cnt[maxp], choice[maxn];
int ksm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return res;
}
void sieve(int n) {
  c[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (c[i] == 0) P[++ps] = i, minn[i] = i;
    for (int j = 1; j <= ps; j++) {
      if (i * P[j] > n) break;
      minn[i * P[j]] = P[j], c[i * P[j]] = 1;
      if (i % P[j] == 0) break;
    }
  }
}
inline void update(int x, int v) {
  if (maxx[x] == v) cnt[x]++;
  if (maxx[x] < v) maxx[x] = v, cnt[x] = 1;
}
int check(int now) {
  while (now > 1) {
    int v = minn[now], c = 0;
    while (now % v == 0) now /= v, c++;
    if (maxx[v] == c && cnt[v] == 1) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + 1 + n), sieve(p[n]);
  lcm = 1;
  for (int i = n; i >= 1; i--) {
    if (maxx[p[i]] == 0) {
      choice[i] = 1;
      update(p[i], 1), lcm = 1ll * lcm * p[i] % mod;
      continue;
    }
    int now = p[i] - 1;
    while (now > 1) {
      int v = minn[now], c = 0;
      while (now % v == 0) now /= v, c++;
      lcm = 1ll * lcm * ksm(v, max(c - maxx[v], 0)) % mod, update(v, c);
    }
  }
  for (int i = 1; i <= n; i++)
    if ((choice[i] == 0 && check(p[i] - 1)) || (choice[i] && check(p[i]))) {
      extra = 1;
      break;
    }
  printf("%d\n", (lcm + extra) % mod);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
const int maxn = 200005, maxp = 2000005, mod = 1000000007;
int n, ps, lcm, extra;
int p[maxn], minn[maxp], P[maxp], c[maxp], maxx[maxp], cnt[maxp], choice[maxn];
int ksm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return res;
}
void sieve(int n) {
  c[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (c[i] == 0) P[++ps] = i, minn[i] = i;
    for (int j = 1; j <= ps; j++) {
      if (i * P[j] > n) break;
      minn[i * P[j]] = P[j], c[i * P[j]] = 1;
      if (i % P[j] == 0) break;
    }
  }
}
inline void update(int x, int v) {
  if (maxx[x] == v) cnt[x]++;
  if (maxx[x] < v) maxx[x] = v, cnt[x] = 1;
}
int check(int now) {
  while (now > 1) {
    int v = minn[now], c = 0;
    while (now % v == 0) now /= v, c++;
    if (maxx[v] == c && cnt[v] == 1) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  sort(p + 1, p + 1 + n), sieve(p[n]);
  lcm = 1;
  for (int i = n; i >= 1; i--) {
    if (maxx[p[i]] == 0) {
      choice[i] = 1;
      update(p[i], 1), lcm = 1ll * lcm * p[i] % mod;
      continue;
    }
    int now = p[i] - 1;
    while (now > 1) {
      int v = minn[now], c = 0;
      while (now % v == 0) now /= v, c++;
      lcm = 1ll * lcm * ksm(v, max(c - maxx[v], 0)) % mod, update(v, c);
    }
  }
  for (int i = 1; i <= n; i++)
    if ((choice[i] == 0 && check(p[i] - 1)) || (choice[i] && check(p[i]))) {
      extra = 1;
      break;
    }
  printf("%d\n", (lcm + extra) % mod);
  return 0;
}

```