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 dp[2000 + 1];
int v[2000 + 1];
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int main() {
  int i, j, N, t = -1, res = 0;
  scanf("%d", &N);
  for (i = 1; i <= N; ++i) scanf("%d", &v[i]);
  for (i = 1; i <= N; ++i)
    if (v[i] != 1) ++res;
  for (i = 1; i <= N; ++i) dp[i] = v[i];
  if (res != N) {
    printf("%d\n", res);
    return 0;
  }
  for (i = 2; i <= N; ++i) {
    for (j = 1; j <= N - i + 1; ++j) {
      dp[j] = gcd(dp[j], v[j + i - 1]);
      if (dp[j] == 1) {
        t = i;
        i = j = N + 1;
      }
    }
  }
  if (t < 0)
    printf("-1\n");
  else
    printf("%d\n", res + t - 2);
  return 0;
}
