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;
int gcd(int a, int b) {
  int t;
  while (b) {
    t = a % b;
    a = b;
    b = t;
  }
  return a;
}
int main() {
  int n;
  int cnt1 = 0;
  scanf("%d", &n);
  vector<int> vals(n);
  vector<vector<int> > dp(n, vector<int>(n, n + 1));
  for (int i = 0; i < n; i++) {
    scanf("%d", &vals[i]);
    dp[i][i] = vals[i];
    cnt1 += vals[i] == 1;
  }
  int g, r;
  int result = -1;
  if (cnt1) {
    result = n - cnt1;
    printf("%d\n", result);
  } else {
    for (int len = 1; len <= n; len++) {
      for (int i = 0; i < n; i++) {
        r = i + len;
        if (r >= n) continue;
        g = dp[i][r] = gcd(dp[i][r - 1], dp[r][r]);
        if (g == 1) break;
      }
      if (g == 1) {
        result = len + n - 1;
        break;
      }
    }
    printf("%d\n", result);
  }
  return 0;
}
