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 MOD = 1E9 + 7;
inline int Add(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int Sub(int a, int b) { return (a >= b) ? (a - b) : (a - b + MOD); }
inline int Mul(int a, int b) { return (long long)a * b % MOD; }
const int N = 200005, P = 2000005;
int n;
int p[P], mnp[P], mnpc[P], mnpp[P];
bool ntp[P];
void Sieve(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!ntp[i]) {
      p[++*p] = i;
      mnp[i] = mnpp[i] = i;
      mnpc[i] = 1;
    }
    for (int j = 1, t = 0; j <= *p && (t = p[j] * i) <= n; ++j) {
      ntp[t] = 1;
      if (i % p[j]) {
        mnp[t] = mnpp[t] = p[j];
        mnpc[t] = 1;
      } else {
        mnp[t] = mnp[i];
        mnpc[t] = mnpc[i] + 1;
        mnpp[t] = mnpp[i] * p[j];
        break;
      }
    }
  }
  return;
}
int cnt[P], r[P], c[P], v[P];
int main() {
  Sieve(P - 1);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  int tp = 0;
  for (int i = P - 1; i; --i)
    if (cnt[i]) {
      if (!c[i]) {
        r[i] = i;
        c[i] = 1;
        if (cnt[i] > 1) {
          v[i] = 1;
          for (int x = i - 1; x > 1; x /= mnpp[x])
            if (r[mnp[x]] < mnpp[x]) {
              r[mnp[x]] = mnpp[x];
              c[mnp[x]] = 1;
            } else if (r[mnp[x]] == mnpp[x])
              c[mnp[x]]++;
        }
        if (cnt[i] > 2) {
          v[i] = 0;
          tp = 1;
        }
      } else {
        v[i] = 1;
        for (int x = i - 1; x > 1; x /= mnpp[x])
          if (r[mnp[x]] < mnpp[x]) {
            r[mnp[x]] = mnpp[x];
            c[mnp[x]] = 1;
          } else if (r[mnp[x]] == mnpp[x])
            c[mnp[x]]++;
        if (cnt[i] > 1) {
          v[i] = 0;
          tp = 1;
        }
      }
    }
  for (int i = 1; i < P && !tp; ++i)
    if (v[i]) {
      bool flag = 1;
      for (int x = i - 1; x > 1 && flag; x /= mnpp[x])
        if (r[mnp[x]] < mnpp[x] || (r[mnp[x]] == mnpp[x] && c[mnp[x]] == 1))
          flag = 0;
      if (flag) tp = 1;
    }
  int ret = 1;
  for (int i = 1; i < P; ++i)
    if (c[i]) ret = Mul(ret, r[i]);
  ret = Add(ret, tp);
  printf("%d\n", ret);
  return 0;
}