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 INF = 0x7fffffff;
const int maxn = 2e4 + 7;
int a[maxn];
long long _gcd(long long a, long long b) { return a == 0 ? b : _gcd(b % a, a); }
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    int minn = INF;
    int flag = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == 1) {
        flag++;
      }
    }
    if (flag) {
      cout << n - flag << endl;
      continue;
    }
    for (int i = 0; i < n; ++i) {
      int gcd = a[i];
      for (int j = i + 1; j < n; j++) {
        gcd = _gcd(gcd, a[j]);
        if (gcd == 1) {
          minn = min(minn, j - i);
          break;
        }
      }
    }
    if (minn == INF) {
      cout << -1 << endl;
    } else {
      cout << minn + n - 1 << endl;
    }
  }
  return 0;
}
