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 n, h;
int dp[2017][2017];
int a[2017];
int gcd(int x, int y) {
  if (x % y == 0) {
    return y;
  } else {
    return gcd(y, x % y);
  }
}
void solve() {
  int i, j, k = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    dp[i][1] = a[i];
    if (a[i] == 1) {
      k++;
    }
  }
  if (k > 0) {
    printf("%d\n", n - k);
    return;
  }
  for (j = 2; j <= n; j++) {
    for (i = 0; i + j <= n; i++) {
      dp[i][j] = gcd(dp[i][j - 1], a[i + j - 1]);
      if (dp[i][j] == 1) {
        printf("%d\n", j - 1 + n - 1);
        return;
      }
    }
  }
  printf("-1\n");
}
int main() {
  while (scanf("%d", &n) != EOF) {
    solve();
  }
  return 0;
}
