You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say x and y, and replace one of them with gcd(x, y), where gcd denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples

Input

5
2 2 3 4 6


Output

5


Input

4
2 4 6 8


Output

-1


Input

3
2 6 9


Output

4

Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

  * [2, 2, 3, 4, 6]. 
  * [2, 1, 3, 4, 6]
  * [2, 1, 3, 1, 6]
  * [2, 1, 1, 1, 6]
  * [1, 1, 1, 1, 6]
  * [1, 1, 1, 1, 1]



We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
#include <bits/stdc++.h>
using namespace std;
constexpr int inf = 0x3f3f3f3f;
class Solver {
 public:
  explicit Solver(vector<int>&& v)
      : n_(v.size()),
        v_(std::move(v)),
        gcd_(n_, vector<int>(n_ + 1, -1)),
        steps_(n_, vector<int>(n_ + 1, -1)) {}
  int solve() { return g(0, n_); }

 private:
  int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int f(int i, int j) {
    if (i + 1 == j) return v_[i];
    int& ans = gcd_[i][j];
    if (ans != -1) return ans;
    return ans = gcd(v_[i], f(i + 1, j));
  }
  int g(int i, int j) {
    if (i + 1 == j) return v_[i] == 1 ? 0 : inf;
    int& ans = steps_[i][j];
    if (ans != -1) return ans;
    ans = inf;
    if (f(i + 1, j) == 1) {
      ans = min(ans, (v_[i] == 1 ? 0 : 1) + g(i + 1, j));
    }
    if (f(i, j - 1) == 1) {
      ans = min(ans, (v_[j - 1] == 1 ? 0 : 1) + g(i, j - 1));
    }
    if (f(i, j) == 1) {
      ans = min(ans, 2 * (j - i) - 2);
    }
    return ans;
  }
  int n_;
  vector<int> v_;
  vector<vector<int>> gcd_;
  vector<vector<int>> steps_;
};
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int& e : v) cin >> e;
  Solver solver(std::move(v));
  int tmp = solver.solve();
  cout << (tmp == inf ? -1 : tmp) << endl;
  return 0;
}
