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>
long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
std::mt19937 rnd(time(NULL));
template <class T>
void cxk(T& a, T b) {
  a = a > b ? a : b;
}
template <class T>
void cnk(T& a, T b) {
  a = a < b ? a : b;
}
int pow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
template <class Ta, class Tb>
void inc(Ta& a, Tb b) {
  a = a + b >= 1000000007 ? a + b - 1000000007 : a + b;
}
template <class Ta, class Tb>
void dec(Ta& a, Tb b) {
  a = a >= b ? a - b : a + 1000000007 - b;
}
int p[200010];
bool yes[2000010];
int d[2000010], pr[200010], P, pd[2000010];
int pw[200010], cnt[200010];
int main() {
  int n = gi();
  for (int i = 2; i <= 2000000; ++i) {
    if (!yes[i]) pr[++P] = i, d[i] = P, pd[i] = 1;
    for (int j = 1; j <= P && i * pr[j] <= 2000000; ++j) {
      yes[i * pr[j]] = 1;
      d[i * pr[j]] = j;
      if (i % pr[j] == 0) {
        pd[i * pr[j]] = pd[i] + 1;
        break;
      }
      pd[i * pr[j]] = 1;
    }
  }
  for (int i = 1; i <= n; ++i)
    p[i] = std::lower_bound(pr + 1, pr + P + 1, gi()) - pr;
  std::sort(p + 1, p + n + 1);
  for (int i = n; i; --i) {
    int x = pr[p[i]];
    if (!pw[p[i]])
      pw[p[i]] = cnt[p[i]] = 1, p[i] = 0;
    else {
      --x;
      while (x > 1) {
        int y = d[x], py = pd[x];
        if (pw[y] < py)
          pw[y] = py, cnt[y] = 1;
        else if (pw[y] == py)
          ++cnt[y];
        while (x % pr[y] == 0) x /= pr[y];
      }
    }
  }
  int qwq = 0;
  for (int i = 1; i <= n; ++i) {
    int x = pr[p[i]] - 1, now = 1;
    while (x > 1) {
      int y = d[x], py = pd[x];
      if (pw[y] == py && cnt[y] == 1) now = 0;
      while (x % pr[y] == 0) x /= pr[y];
    }
    if (x == 1) qwq |= now;
  }
  int ans = 1;
  for (int i = 1; i <= P; ++i) ans = 1ll * ans * pow(pr[i], pw[i]) % 1000000007;
  printf("%d\n", (ans + qwq) % 1000000007);
  return 0;
}