#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200100;
const int MAXP = 2000100;
const long long MOD = 1e9 + 7;
int N;
int pprime[MAXP];
int arr[MAXN];
int pord[MAXN];
map<int, int> pv;
map<int, int> nd;
int nvp[MAXP];
int nf[MAXP];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> N;
  for (int i = 2; i < MAXP; i++) {
    if (!pprime[i]) {
      pprime[i] = i;
      for (int j = i; j < MAXP; j += i) pprime[j] = i;
    }
  }
  for (int i = 0; i < N; i++) cin >> arr[i];
  sort(arr, arr + N);
  reverse(arr, arr + N);
  int r = 0;
  for (int i = 0; i < N; i++) {
    int cp = arr[i];
    bool gadd = false;
    if (pv.find(cp) != pv.end()) {
      cp--;
      while (cp > 1) {
        int np = pprime[cp], nc = 0;
        while (cp % np == 0) {
          cp /= np;
          nc++;
        }
        pv[np];
        if (pv[np] < nc) {
          pv[np] = nc;
          nd[np] = 1;
          gadd = true;
        } else if (pv[np] == nc)
          nd[np]++;
      }
      pord[i] = arr[i] - 1;
    } else {
      pv[cp] = 1;
      nd[cp] = 1;
      gadd = true;
      pord[i] = arr[i];
    }
    if (!gadd) r = 1;
  }
  for (pair<int, int> ppair : pv) {
    nvp[ppair.first] = ppair.second;
  }
  for (int i = 0; i < N; i++) {
    int cp = pord[i];
    bool gadd = false;
    while (cp > 1) {
      int np = pprime[cp], nc = 0;
      while (cp % np == 0) {
        cp /= np;
        nc++;
      }
      if (pv[np] == nc && nd[np] == 1) gadd = true;
    }
    if (!gadd) r = 1;
  }
  long long res = 1;
  for (pair<int, int> ppair : pv) {
    for (int i = 0; i < ppair.second; i++) res = (res * ppair.first) % MOD;
  }
  cout << (res + r) % MOD << "\n";
}

### Prompt

In Cpp, your task is to solve the following 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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200100;
const int MAXP = 2000100;
const long long MOD = 1e9 + 7;
int N;
int pprime[MAXP];
int arr[MAXN];
int pord[MAXN];
map<int, int> pv;
map<int, int> nd;
int nvp[MAXP];
int nf[MAXP];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> N;
  for (int i = 2; i < MAXP; i++) {
    if (!pprime[i]) {
      pprime[i] = i;
      for (int j = i; j < MAXP; j += i) pprime[j] = i;
    }
  }
  for (int i = 0; i < N; i++) cin >> arr[i];
  sort(arr, arr + N);
  reverse(arr, arr + N);
  int r = 0;
  for (int i = 0; i < N; i++) {
    int cp = arr[i];
    bool gadd = false;
    if (pv.find(cp) != pv.end()) {
      cp--;
      while (cp > 1) {
        int np = pprime[cp], nc = 0;
        while (cp % np == 0) {
          cp /= np;
          nc++;
        }
        pv[np];
        if (pv[np] < nc) {
          pv[np] = nc;
          nd[np] = 1;
          gadd = true;
        } else if (pv[np] == nc)
          nd[np]++;
      }
      pord[i] = arr[i] - 1;
    } else {
      pv[cp] = 1;
      nd[cp] = 1;
      gadd = true;
      pord[i] = arr[i];
    }
    if (!gadd) r = 1;
  }
  for (pair<int, int> ppair : pv) {
    nvp[ppair.first] = ppair.second;
  }
  for (int i = 0; i < N; i++) {
    int cp = pord[i];
    bool gadd = false;
    while (cp > 1) {
      int np = pprime[cp], nc = 0;
      while (cp % np == 0) {
        cp /= np;
        nc++;
      }
      if (pv[np] == nc && nd[np] == 1) gadd = true;
    }
    if (!gadd) r = 1;
  }
  long long res = 1;
  for (pair<int, int> ppair : pv) {
    for (int i = 0; i < ppair.second; i++) res = (res * ppair.first) % MOD;
  }
  cout << (res + r) % MOD << "\n";
}

```