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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
const int mod = 1000000007, MAX = 200005, INF = 1 << 30;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  std::ifstream in("text.txt");
  std::cin.rdbuf(in.rdbuf());
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<long long> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];
  bool aru = false;
  for (int i = 0; i < N; i++)
    if (A[i] == 1) aru = true;
  if (aru) {
    int ans = 0;
    for (int i = 0; i < N; i++)
      if (A[i] != 1) ans++;
    cout << ans << endl;
    return 0;
  }
  int need = INF;
  for (int i = 0; i < N; i++) {
    long long g = 0;
    for (int j = i; j < N; j++) {
      g = gcd(g, A[j]);
      if (g == 1) {
        chmin(need, j - i + 1);
        break;
      }
    }
  }
  if (need == INF)
    cout << -1 << endl;
  else
    cout << need - 1 + N - 1 << endl;
}
