#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5 + 1, maxm = (int)2e6 + 1, mod = (int)1e9 + 7;
int tot, pr[maxm], d[maxm];
int n, m, a[maxn];
pair<int, int> mx[maxm];
int main() {
  d[1] = 1;
  for (int i = 2; i < maxm; ++i) {
    if (!d[i]) pr[tot++] = d[i] = i;
    for (int j = 0, k; (k = i * pr[j]) < maxm; ++j) {
      d[k] = pr[j];
      if (d[i] == pr[j]) break;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  sort(a, a + n);
  for (int i = n - 1; i >= 0; --i)
    if (!mx[a[i]].first) {
      mx[a[i]] = make_pair(a[i], 1);
    } else {
      --a[i];
      for (int tmp = a[i]; tmp > 1;) {
        int p = d[tmp], pp = 1;
        for (; d[tmp] == p; tmp /= p, pp *= p)
          ;
        if (mx[p].first < pp) {
          mx[p] = make_pair(pp, 1);
        } else if (mx[p].first == pp) {
          ++mx[p].second;
        }
      }
    }
  bool adt = 0;
  for (int i = 0; !adt && i < n; ++i) {
    bool chk = 0;
    for (int tmp = a[i]; !chk && tmp > 1;) {
      int p = d[tmp], pp = 1;
      for (; d[tmp] == p; tmp /= p, pp *= p)
        ;
      chk |= mx[p].first == pp && mx[p].second == 1;
    }
    adt |= !chk;
  }
  int ans = 1;
  for (int i = 0; i < tot; ++i)
    if (mx[pr[i]].first) ans = (long long)ans * mx[pr[i]].first % mod;
  if (adt) (++ans) >= mod && (ans -= mod);
  printf("%d\n", ans);
  return 0;
}

### 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>
using namespace std;
const int maxn = (int)2e5 + 1, maxm = (int)2e6 + 1, mod = (int)1e9 + 7;
int tot, pr[maxm], d[maxm];
int n, m, a[maxn];
pair<int, int> mx[maxm];
int main() {
  d[1] = 1;
  for (int i = 2; i < maxm; ++i) {
    if (!d[i]) pr[tot++] = d[i] = i;
    for (int j = 0, k; (k = i * pr[j]) < maxm; ++j) {
      d[k] = pr[j];
      if (d[i] == pr[j]) break;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  sort(a, a + n);
  for (int i = n - 1; i >= 0; --i)
    if (!mx[a[i]].first) {
      mx[a[i]] = make_pair(a[i], 1);
    } else {
      --a[i];
      for (int tmp = a[i]; tmp > 1;) {
        int p = d[tmp], pp = 1;
        for (; d[tmp] == p; tmp /= p, pp *= p)
          ;
        if (mx[p].first < pp) {
          mx[p] = make_pair(pp, 1);
        } else if (mx[p].first == pp) {
          ++mx[p].second;
        }
      }
    }
  bool adt = 0;
  for (int i = 0; !adt && i < n; ++i) {
    bool chk = 0;
    for (int tmp = a[i]; !chk && tmp > 1;) {
      int p = d[tmp], pp = 1;
      for (; d[tmp] == p; tmp /= p, pp *= p)
        ;
      chk |= mx[p].first == pp && mx[p].second == 1;
    }
    adt |= !chk;
  }
  int ans = 1;
  for (int i = 0; i < tot; ++i)
    if (mx[pr[i]].first) ans = (long long)ans * mx[pr[i]].first % mod;
  if (adt) (++ans) >= mod && (ans -= mod);
  printf("%d\n", ans);
  return 0;
}

```