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].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int RLEN = 1 << 18 | 1;
inline char nc() {
  static char ibuf[RLEN], *ib, *ob;
  (ib == ob) && (ob = (ib = ibuf) + fread(ibuf, 1, RLEN, stdin));
  return (ib == ob) ? -1 : *ib++;
}
inline int rd() {
  char ch = nc();
  int i = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = nc();
  }
  while (isdigit(ch)) {
    i = (i << 1) + (i << 3) + ch - '0';
    ch = nc();
  }
  return i * f;
}
const int N = 2e6 + 50, mod = 1e9 + 7;
int pr[N], pt, npr[N], mnpr[N];
int n, p[N], c[N], mx[N];
inline void sieve() {
  for (int i = 2; i < N; i++) {
    if (!npr[i]) pr[++pt] = i, mnpr[i] = i;
    for (int j = 1; j <= pt; j++) {
      long long k = i * pr[j];
      if (k >= N) break;
      mnpr[k] = pr[j];
      npr[k] = 1;
      if (!(i % pr[j])) break;
    }
  }
}
int main() {
  sieve();
  n = rd();
  for (int i = 1; i <= n; i++) p[i] = rd();
  sort(p + 1, p + n + 1);
  int bz = 0, ans = 1;
  for (int i = n; i >= 1; i--) {
    if (!c[p[i]])
      ++c[p[i]], ans = (long long)ans * p[i] % mod, mx[p[i]] = 1;
    else {
      int v = --p[i];
      while (v != 1) {
        int u = mnpr[v], ci = 0;
        while (!(v % u)) v /= u, ++ci;
        if (c[u] < ci) {
          for (int j = c[u] + 1; j <= ci; j++) ans = (long long)ans * u % mod;
          c[u] = ci;
          mx[u] = 1;
        } else
          mx[u] += (c[u] == ci);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int v = p[i], tag = 0;
    while (v != 1) {
      int u = mnpr[v], ci = 0;
      while (!(v % u)) v /= u, ++ci;
      if (c[u] == ci && mx[u] == 1) tag = 1;
    }
    if (!tag) {
      ++ans;
      break;
    }
  }
  cout << ans << '\n';
}