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[2010][2010], a[2010];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, i, j, ans, mn, cnt;
  scanf("%d", &n);
  cnt = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1) cnt++;
  }
  if (cnt) {
    ans = n - cnt;
  } else {
    mn = 0x3fffffff;
    for (i = 0; i < n; i++) {
      for (j = i; j < n; j++) {
        dp[i][j] = (i == j ? a[i] : gcd(dp[i][j - 1], a[j]));
        if (dp[i][j] == 1) mn = min(mn, j - i);
      }
    }
    if (mn == 0x3fffffff)
      ans = -1;
    else
      ans = mn + n - 1;
  }
  printf("%d\n", ans);
  return 0;
}
