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 long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const int N = 2e6 + 100, M = 1e9 + 7;
bool mark[N];
int prime[N];
void crivo() {
  for (int i = 2; i < N; i++) {
    if (!mark[i]) {
      prime[i] = i;
      for (int j = i + i; j < N; j += i) {
        mark[j] = 1;
        prime[j] = i;
      }
    }
  }
}
int d[N];
bool some(int x) {
  while (x > 1) {
    int p = prime[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    if (d[p] < k) {
      return true;
    }
  }
  return false;
}
multiset<int> S[N];
int usado[N];
void att(long long x) {
  usado[x] = 1;
  while (x > 1) {
    int p = prime[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    d[p] = max(d[p], k);
    S[p].insert(k);
  }
}
bool bad(int x) {
  bool ok = 1;
  int xx = x;
  while (x > 1) {
    int p = prime[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    S[p].erase(S[p].find(k));
    if (S[p].size() > 0 and *S[p].rbegin() >= d[p]) {
    } else {
      S[p].insert(k);
      return false;
    }
    S[p].insert(k);
  }
  return true;
}
long long lcm() {
  long long ans = 1;
  for (int i = 1; i < N; i++) {
    if (d[i] == 0) continue;
    ans = (ans * exp(i, d[i], M)) % M;
  }
  return ans;
}
int p[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  crivo();
  int n;
  cin >> n;
  int pp = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
  }
  sort(p + 1, p + 1 + n);
  for (int i = n; i >= 1; i--) {
    if (some(p[i])) {
      att(p[i]);
    } else if (some(p[i] - 1)) {
      att(p[i] - 1);
    } else
      pp = 1;
  }
  for (int i = 1; i < N; i++) {
    if (pp) break;
    if (usado[i] and bad(i)) {
      pp = 1;
      break;
    }
  }
  long long ans = lcm();
  ans = mod(ans + pp, M);
  cout << ans << "\n";
}