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;
inline int read() {
  int k = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  return k * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(int x) {
  write(x);
  puts("");
}
const int N = 2010;
int n, a[N];
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  n = read();
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i] == 1) cnt++;
  }
  int ans = 1e9;
  for (int i = 1; i <= n; i++) {
    int p = a[i];
    for (int j = i; j <= n; j++) {
      p = gcd(p, a[j]);
      if (p == 1) ans = min(ans, j - i);
    }
  }
  cnt = max(cnt, 1);
  if (ans < 1e9)
    writeln(ans + (n - cnt));
  else
    puts("-1");
  return 0;
}
