codeforces 891-A. Pride Codeforces Round #446 (Div. 2) C. Pride

C. Pride

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say xand 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, ..., 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.

思路：

先求出总的区间的gcd，如果不是1，输出-1；如果是1，那么我们还要考虑1的出现次数和最短的gcd=1的区间长度（将这段区间先造出1）。

AC code

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;

ll  res[2005];

int qgcd(int a,int b)
{
    return b==0?a:qgcd(b,a%b);
}

int main()
{
    int n;scanf("%d",&n);
    int gcd,sum=0;
    scanf("%d",&res[1]);
    gcd=res[1];
    if(gcd==1)
        sum++;
    for(int i=2;i<=n;i++)
    {
        scanf("%d",&res[i]);
        if(res[i]==1)
            sum++;
        gcd=qgcd(gcd,res[i]);
    }
    if(gcd!=1)
    {
        cout<<-1<<endl;
        return 0;
    }
    else
    {
        if(sum!=0)
        {
            cout<<n-sum<<endl;
            return 0;
        }
        int minn=n;
        for(int i=1; i<=n; i++)
        {
            gcd=res[i];
            for(int j=i+1; j<=n; j++)
            {
                gcd=qgcd(gcd,res[j]);
                if(gcd==1)
                {
                    minn=min(minn,j-i+1);
                    break;
                }
            }
        }
        cout<<n+minn-2<<endl;
    }
    return 0;
}