CodeForces - 892C Pride

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.

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, ..., a_n (1 ≤ a_i ≤ 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.

Example

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<stdio.h>
int n;
int a[2200];
int gcd(int a,int b)
{
    if(b==0)return a;
    else return gcd(b,a%b);
}
int main()
{
    while(~scanf("%d",&n))
    {
        int f=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==1)
                f++;
        }
        if(f)//有1的话，就很好处理了。
        {
            printf("%d\n",n-f);
            continue;
        }
        int ans=-1;
        int t=n;
        for(int i=1; i<=n; i++)//进行n次循环
        {
            for(int j=1; j<t; j++)  //缩小范围
            {
                a[j]=gcd(a[j],a[j+1]);//把公约数赋给a[j];，a[j+1]才可以不用
                if(a[j]==1)//出现1后就可以跳出循环了
                {
                    ans=i;
                    break;
                }
            }
            if(ans!=-1)
            {
                ans=ans+n-1;
                break;
            }
            t--;
        }
        printf("%d\n",ans);
    }
    return 0;
}