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 md = 1e9 + 7;
bool cmax(int &x, int y) { return x < y ? x = y, true : false; }
int mul(int x, int y) { return (long long)x * y % md; }
int power(int x, int y) {
  int result = 1;
  for (; y; y >>= 1, x = mul(x, x)) {
    if (y & 1) {
      result = mul(result, x);
    }
  }
  return result;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a.begin(), a.end());
  int m = a[n - 1];
  vector<int> prime, minp(m + 1);
  for (int i = 2; i <= m; ++i) {
    if (!minp[i]) {
      prime.push_back(i);
      minp[i] = i;
    }
    for (auto p : prime) {
      if (i * p > m) {
        break;
      }
      minp[i * p] = p;
      if (i % p == 0) {
        break;
      }
    }
  }
  vector<int> appear(m + 1);
  for (auto x : a) {
    ++appear[x];
  }
  bool add = false;
  vector<int> cover(m + 1), times(m + 1);
  for (int i = 1; i <= m; ++i) {
    if (appear[i] == 3) {
      add = true;
    }
  }
  vector<int> all;
  auto insert = [&](int x) {
    bool useful = false;
    int old = x;
    while (x != 1) {
      int p = minp[x], e = 0;
      while (x % p == 0) {
        x /= p;
        ++e;
      }
      if (cmax(cover[p], e)) {
        useful = true;
        times[p] = 1;
      } else if (cover[p] == e) {
        ++times[p];
      }
    }
    if (useful) {
      all.push_back(old);
    } else {
      x = old;
      while (x != 1) {
        int p = minp[x], e = 0;
        while (x % p == 0) {
          x /= p;
          ++e;
        }
        if (cover[p] == e) {
          --times[p];
        }
      }
    }
    return useful;
  };
  for (int i = m; i; --i) {
    if (appear[i] >= 2) {
      if (!insert(i)) {
        add = true;
      }
      if (!insert(i - 1)) {
        add = true;
      }
    }
  }
  for (int i = m; i; --i) {
    if (appear[i] == 1) {
      if (!insert(i)) {
        if (!insert(i - 1)) {
          add = true;
        }
      }
    }
  }
  auto check = [&](int x) {
    bool useful = false;
    while (x != 1) {
      int p = minp[x], e = 0;
      while (x % p == 0) {
        x /= p;
        ++e;
      }
      if (cover[p] == e && times[p] == 1) {
        useful = true;
        break;
      }
    }
    return !useful;
  };
  for (auto x : all) {
    if (check(x)) {
      add = true;
      break;
    }
  }
  int answer = 1;
  for (int i = 2; i <= m; ++i) {
    answer = mul(answer, power(i, cover[i]));
  }
  if (add) {
    answer = (answer + 1) % md;
  }
  printf("%d\n", answer);
  return 0;
}