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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 2000006;
int freq[N];
int g[N];
map<int, int> ms[N];
int mx[N];
int factor[N];
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
vector<pair<int, int> > changes;
int ans = 1;
int getChange(int x) {
  int ret = 1;
  changes.clear();
  while (x != 1) {
    int p = factor[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    if (mx[p] < k) {
      changes.push_back({p, mx[p]});
      ret = mul(ret, powr(p, k - mx[p]));
      mx[p] = k;
    }
  }
  reverse(changes.begin(), changes.end());
  for (auto it : changes) mx[it.first] = it.second;
  return ret;
}
void add(int x) {
  while (x != 1) {
    int p = factor[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    if (mx[p] < k) {
      ans = mul(ans, powr(p, k - mx[p]));
      mx[p] = k;
    }
    ms[p][k]++;
  }
}
bool canRemove(int x) {
  while (x != 1) {
    int p = factor[x];
    int k = 0;
    while (x % p == 0) {
      x /= p;
      k++;
    }
    if (mx[p] == k) {
      if (ms[p][k] == 1) return 0;
    }
  }
  return 1;
}
bool P[N], Q[N];
bool R[N];
int main() {
  int n;
  scanf("%d", &(n));
  for (int i = 2; i < N; i++) {
    if (!factor[i])
      for (int j = i; j < N; j += i) factor[j] = i;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &(x));
    freq[x]++;
  }
  for (int i = N - 1; i >= 2; i--) {
    if (freq[i] >= 2) {
      add(i);
      add(i - 1);
      P[i] = Q[i] = 1;
    } else if (freq[i] == 1) {
      int x1 = getChange(i), x2 = getChange(i - 1);
      if (x1 > x2) {
        add(i);
        P[i] = 1;
      } else {
        add(i - 1);
        Q[i] = 1;
      }
    }
    if (freq[i] > 2) R[i] = 1;
  }
  for (int i = 2; i < N; i++) {
    if (R[i]) {
      ans = add(ans, 1);
      printf("%d\n", ans);
      return 0;
    }
    if (P[i]) {
      if (canRemove(i)) {
        ans = add(ans, 1);
        printf("%d\n", ans);
        return 0;
      }
    }
    if (Q[i]) {
      if (canRemove(i - 1)) {
        ans = add(ans, 1);
        printf("%d\n", ans);
        return 0;
      }
    }
  }
  printf("%d\n", ans);
}