Pride （思维）

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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.

Examples

Input

5
2 2 3 4 6

Output

Input

4
2 4 6 8

Output

-1

Input

3
2 6 9

Output

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.

题意：你在给的数字里面选两个数，可以用这两数的最大公约数来替换其中一个，问最少几次可以他们都替换成1，如果无法做成，输出-1，否则输出替换次数。

思路其实可以分两类，如果这n个数字里面含有m个1，那么交换次数就是n-m，因为只要有1，就可以把另一个任意数字换成1，如果没有1，那我们需要求一个最小的把任意一个书换成1的次数m，然后用n-1+m（用m次交换把数组变成了只有一个1的数组，那么有了1个1还有n-1个需要变所以总次数就是n-1+m）。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int gcd(int a,int b){
    if(b == 0) return a;
    else return gcd(b,a%b);
}
int main(){
    int n;
    int a[5200];
    scanf("%d",&n);
    int cnt = 0;
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        if(a[i] == 1) cnt++;
    }
    if(cnt){
        cout << n-cnt << endl;
        return 0;
    }
    int mint = 99999999;
    for(int i = 1; i <= n; i++){
        int x = a[i];
        cnt = 0;
        for(int j = i+1; j <= n; j++){
            cnt++;
            x = gcd(x,a[j]);
            if(x == 1) break;
        }
        if(x == 1)
            mint = min(mint,cnt);
    }
    if(mint == 99999999)
        cout << "-1" << endl;
    else
        cout << n-1+mint << endl;
    return 0;
}