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 x, int y) {
  int t;
  while (y) {
    t = x % y;
    x = y;
    y = t;
  }
  return x;
}
int dp[2005][2005];
int main() {
  int n, flag, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &dp[i][i]), ans += (dp[i][i] != 1);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) dp[i][j] = gcd(dp[i][j - 1], dp[j][j]);
  for (int s = 1; s <= n; s++) {
    flag = 0;
    for (int i = 1, cur; (cur = i + s - 1) <= n; i++) {
      if (dp[i][cur] == 1) {
        if (s == 1)
          printf("%d", ans);
        else
          printf("%d", n + s - 2);
        return 0;
      }
    }
  }
  printf("-1");
}
