#include <bits/stdc++.h>
using namespace std;
const int MOD = int(1e9) + 7;
int mul(int a, int b) { return int(a * 1ll * b % MOD); }
int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
const int M = 2000 * 1000 + 555;
int minD[M];
vector<pair<int, int> > ps[M];
inline void precalc() {
  iota(minD, minD + M, 0);
  for (int val = int(2); val < int(M); val++) {
    if (minD[val] < val) continue;
    for (long long j = val * 1ll * val; j < M; j += val)
      minD[j] = min(minD[j], val);
  }
  for (int i = int(2); i < int(M); i++) {
    int v = i;
    ps[i].clear();
    while (v > 1) {
      int md = minD[v];
      if (ps[i].empty() || ps[i].back().first != md) ps[i].emplace_back(md, 0);
      ps[i].back().second++;
      v /= md;
    }
  }
}
int n;
vector<int> p;
inline bool read() {
  if (!(cin >> n)) return false;
  p.assign(n, 0);
  for (int i = int(0); i < int(n); i++) assert(scanf("%d", &p[i]) == 1);
  return true;
}
int a[M], cnt[M];
inline void solve() {
  sort(p.begin(), p.end(), greater<int>());
  for (int& cp : p) {
    if (a[cp] == 0) {
      a[cp] = 1;
      cnt[cp] = 1;
    } else {
      cp--;
      for (auto np : ps[cp]) {
        if (a[np.first] < np.second) {
          a[np.first] = np.second;
          cnt[np.first] = 1;
        } else if (a[np.first] == np.second)
          cnt[np.first]++;
      }
    }
  }
  int add = 0;
  for (int cp : p) {
    bool un = false;
    for (auto np : ps[cp])
      un |= (a[np.first] == np.second && cnt[np.first] == 1);
    if (!un) add = 1;
  }
  int lcm = 1;
  for (int i = int(2); i < int(M); i++) lcm = mul(lcm, binPow(i, a[i]));
  cout << (lcm + add) % MOD << endl;
}
int main() {
  cout << fixed << setprecision(10);
  precalc();
  if (read()) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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 MOD = int(1e9) + 7;
int mul(int a, int b) { return int(a * 1ll * b % MOD); }
int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
const int M = 2000 * 1000 + 555;
int minD[M];
vector<pair<int, int> > ps[M];
inline void precalc() {
  iota(minD, minD + M, 0);
  for (int val = int(2); val < int(M); val++) {
    if (minD[val] < val) continue;
    for (long long j = val * 1ll * val; j < M; j += val)
      minD[j] = min(minD[j], val);
  }
  for (int i = int(2); i < int(M); i++) {
    int v = i;
    ps[i].clear();
    while (v > 1) {
      int md = minD[v];
      if (ps[i].empty() || ps[i].back().first != md) ps[i].emplace_back(md, 0);
      ps[i].back().second++;
      v /= md;
    }
  }
}
int n;
vector<int> p;
inline bool read() {
  if (!(cin >> n)) return false;
  p.assign(n, 0);
  for (int i = int(0); i < int(n); i++) assert(scanf("%d", &p[i]) == 1);
  return true;
}
int a[M], cnt[M];
inline void solve() {
  sort(p.begin(), p.end(), greater<int>());
  for (int& cp : p) {
    if (a[cp] == 0) {
      a[cp] = 1;
      cnt[cp] = 1;
    } else {
      cp--;
      for (auto np : ps[cp]) {
        if (a[np.first] < np.second) {
          a[np.first] = np.second;
          cnt[np.first] = 1;
        } else if (a[np.first] == np.second)
          cnt[np.first]++;
      }
    }
  }
  int add = 0;
  for (int cp : p) {
    bool un = false;
    for (auto np : ps[cp])
      un |= (a[np.first] == np.second && cnt[np.first] == 1);
    if (!un) add = 1;
  }
  int lcm = 1;
  for (int i = int(2); i < int(M); i++) lcm = mul(lcm, binPow(i, a[i]));
  cout << (lcm + add) % MOD << endl;
}
int main() {
  cout << fixed << setprecision(10);
  precalc();
  if (read()) {
    solve();
  }
  return 0;
}

```