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;
inline void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
const int mod = 1e9 + 7;
int n, m, i, j, a[200005], np[2000005], p[2000005], pw[2000005], wy[2000005],
    mp[2000005], pc, s0, s1;
void push(int x) {
  int i;
  for (((i)) = (1); ((i)) <= ((pc)); ((i))++) {
    if (p[i] * p[i] > x) break;
    if (x % p[i] == 0) {
      int t = 0;
      while (x % p[i] == 0) {
        t++;
        x /= p[i];
      }
      if (t > pw[i]) {
        pw[i] = t;
        wy[i] = 1;
      } else if (t == pw[i]) {
        wy[i]++;
      }
    }
  }
  if (x == 1) return;
  int t = 1;
  i = mp[x];
  if (t > pw[i]) {
    pw[i] = t;
    wy[i] = 1;
  } else if (t == pw[i]) {
    wy[i]++;
  }
}
bool check(int x) {
  int i;
  for (((i)) = (1); ((i)) <= ((pc)); ((i))++) {
    if (p[i] * p[i] > x) break;
    if (x % p[i] == 0) {
      int t = 0;
      while (x % p[i] == 0) {
        t++;
        x /= p[i];
      }
      if (pw[i] < t) return 0;
    }
  }
  if (x == 1) return 1;
  int t = 1;
  i = mp[x];
  if (pw[i] < t) return 0;
  return 1;
}
bool check2(int x) {
  int i;
  for (((i)) = (1); ((i)) <= ((pc)); ((i))++) {
    if (p[i] * p[i] > x) break;
    if (x % p[i] == 0) {
      int t = 0;
      while (x % p[i] == 0) {
        t++;
        x /= p[i];
      }
      if (pw[i] == t && wy[i] == 1) return 0;
    }
  }
  if (x == 1) return 1;
  int t = 1;
  i = mp[x];
  if (pw[i] == t && wy[i] == 1) return 0;
  return 1;
}
int main() {
  for ((i) = (2); (i) <= (2000000); (i)++) {
    if (!np[i]) {
      p[++pc] = i;
      mp[i] = pc;
    }
    for (((j)) = (1); ((j)) <= ((pc)); ((j))++) {
      if (i * p[j] > 2000000) break;
      np[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  read(n);
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) read(a[i]);
  sort(a + 1, a + n + 1);
  reverse(a + 1, a + n + 1);
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
    if (!check(a[i])) {
      push(a[i]);
    } else {
      a[i]--;
      if (!check(a[i])) {
        push(a[i]);
      } else {
        a[i] = 0;
        s1 = 1;
      }
    }
  }
  s0 = 1;
  for (((i)) = (1); ((i)) <= ((pc)); ((i))++) {
    for (((j)) = (1); ((j)) <= ((pw[i])); ((j))++) s0 = 1ll * s0 * p[i] % mod;
  }
  for (((i)) = (1); ((i)) <= ((n)); ((i))++)
    if (a[i] && check2(a[i])) s1 = 1;
  cout << (s0 + s1) % mod << endl;
  return 0;
}