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;
const int M = 100005;
int n, m, x, y;
long long a[2002];
long long sum, k;
string s, z;
vector<pair<int, int> > v;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  cin >> n;
  bool f = true;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] != 1) f = false;
  }
  if (f) {
    cout << 0;
    return 0;
  }
  if (n == 1) {
    if (a[0] == 1)
      cout << 0;
    else
      cout << -1;
    return 0;
  }
  int k = gcd(a[0], a[1]);
  for (int i = 2; i < n; i++) {
    k = gcd(k, a[i]);
  }
  if (k != 1) {
    cout << -1;
    return 0;
  }
  int c = 0, o = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) o++;
  }
  if (o) {
    cout << n - o;
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if (gcd(a[i], a[i + 1]) == 1) {
      cout << n;
      return 0;
    }
  }
  int mn = 2e9;
  for (int i = 0; i < n - 1; i++) {
    k = gcd(a[i], a[i + 1]);
    int temp = k;
    for (int j = i + 1; j < n; j++) {
      k = gcd(k, a[j]);
      if (k == 1) {
        mn = min(mn, j - i);
        break;
      }
    }
    for (int j = i - 1; j >= 0; j--) {
      temp = gcd(temp, a[j]);
      if (temp == 1) {
        mn = min(mn, i - j);
        break;
      }
    }
  }
  cout << n + mn;
  return 0;
}
