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 N = 200010;
const int M = 2000010;
const int MOD = 1000000007;
inline int add(int u, int v) {
  u += v;
  if (u >= MOD) {
    u -= MOD;
  }
  return u;
}
inline int sub(int u, int v) {
  u -= v;
  if (u < 0) {
    u += MOD;
  }
  return u;
}
inline int mul(int u, int v) { return (long long)u * v % MOD; }
inline int power(int u, int v) {
  int res = 1;
  while (v) {
    if (v & 1) {
      res = mul(res, u);
    }
    u = mul(u, u);
    v >>= 1;
  }
  return res;
}
inline int inv(int u) { return power(u, MOD - 2); }
int n;
int p[M];
int a[N];
int maxPower[M], cntMaxPower[M];
vector<pair<int, int> > getPrimeDivs(int u) {
  vector<pair<int, int> > res;
  while (u > 1) {
    int v = p[u];
    res.push_back(make_pair(v, 0));
    while (u % v == 0) {
      u /= v;
      res.back().second++;
    }
  }
  return res;
}
bool canAdd(int u) {
  vector<pair<int, int> > primeDivs = getPrimeDivs(u);
  for (auto v : primeDivs) {
    if (v.second > maxPower[v.first]) {
      return true;
    }
  }
  return false;
}
void add(int u) {
  vector<pair<int, int> > primeDivs = getPrimeDivs(u);
  for (auto v : primeDivs) {
    if (v.second == maxPower[v.first]) {
      cntMaxPower[v.first]++;
    } else if (v.second > maxPower[v.first]) {
      maxPower[v.first] = v.second;
      cntMaxPower[v.first] = 1;
    }
  }
}
int used[N];
bool canRemove(int u) {
  vector<pair<int, int> > primeDivs = getPrimeDivs(u);
  for (auto v : primeDivs) {
    if (v.second == maxPower[v.first] && cntMaxPower[v.first] == 1) {
      return false;
    }
  }
  return true;
}
void remove(int u) {
  vector<pair<int, int> > primeDivs = getPrimeDivs(u);
  for (auto v : primeDivs) {
    if (v.second == maxPower[v.first]) {
      cntMaxPower[v.first]--;
    }
  }
}
int main() {
  for (int i = 2; i < M; i++) {
    if (!p[i]) {
      for (int j = i; j < M; j += i) {
        if (!p[j]) {
          p[j] = i;
        }
      }
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  int tail = 0;
  for (int i = n; i > 0; i--) {
    if (canAdd(a[i])) {
      add(a[i]);
      used[i] = 1;
    } else if (canAdd(a[i] - 1)) {
      used[i] = 2;
      add(a[i] - 1);
    } else {
      tail = 1;
    }
  }
  for (int i = n; i > 0; i--) {
    if (used[i] == 0) {
      continue;
    }
    int u = used[i] == 1 ? a[i] : a[i] - 1;
    if (canRemove(u)) {
      remove(u);
      tail = 1;
    }
  }
  int res = 1;
  for (int i = 1; i < M; i++) {
    if (p[i] == i) {
      res = mul(res, power(i, maxPower[i]));
    }
  }
  res = add(res, tail);
  cout << res << endl;
  return 0;
}