CodeForces - 978D Almost Arithmetic Progression

Polycarp likes arithmetic progressions. A sequence [a1,a2,…,an][a1,a2,…,an] is called an arithmetic progression if for each ii (1≤i<n1≤i<n) the value ai+1−aiai+1−ai is the same. For example, the sequences [42][42], [5,5,5][5,5,5], [2,11,20,29][2,11,20,29] and [3,2,1,0][3,2,1,0] are arithmetic progressions, but [1,0,1][1,0,1], [1,3,9][1,3,9] and [2,3,1][2,3,1] are not.

It follows from the definition that any sequence of length one or two is an arithmetic progression.

Polycarp found some sequence of positive integers [b1,b2,…,bn][b1,b2,…,bn]. He agrees to change each element by at most one. In the other words, for each element there are exactly three options: an element can be decreased by 11, an element can be increased by 11, an element can be left unchanged.

Determine a minimum possible number of elements in bb which can be changed (by exactly one), so that the sequence bb becomes an arithmetic progression, or report that it is impossible.

It is possible that the resulting sequence contains element equals 00.

Input

The first line contains a single integer nn (1≤n≤100000)(1≤n≤100000) — the number of elements in bb.

The second line contains a sequence b1,b2,…,bnb1,b2,…,bn (1≤bi≤109)(1≤bi≤109).

Output

If it is impossible to make an arithmetic progression with described operations, print -1. In the other case, print non-negative integer — the minimum number of elements to change to make the given sequence becomes an arithmetic progression. The only allowed operation is to add/to subtract one from an element (can't use operation twice to the same position).

Examples

Input

4
24 21 14 10

Output

3

Input

2
500 500

Output

0

Input

3
14 5 1

Output

-1

Input

5
1 3 6 9 12

Output

1

Note

In the first example Polycarp should increase the first number on 11, decrease the second number on 11, increase the third number on 11, and the fourth number should left unchanged. So, after Polycarp changed three elements by one, his sequence became equals to [25,20,15,10][25,20,15,10], which is an arithmetic progression.

In the second example Polycarp should not change anything, because his sequence is an arithmetic progression.

In the third example it is impossible to make an arithmetic progression.

In the fourth example Polycarp should change only the first element, he should decrease it on one. After that his sequence will looks like [0,3,6,9,12][0,3,6,9,12], which is an arithmetic progression.

给你一个数列，问能不能仅通过对某个数进行加一减一或不动将他凑成等差数列，可以的话输出操作次数，不行的话输出-1。

首先要明确，我们要找的是构造等差数列看是否符合，而不是一个个数字去试能否构造出来。等差数列的要素有两个：首项和公差。公差可以利用首项和第二项推算出来。也就是说，确定了首项和第二项就能确定一个等差数列。首项有三种变换方式，第二项一样，所以一共可以构造九个等差数列，依次比较是否符合即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long LL;
const int INF=0x3f3f3f3f;

LL a[100010],b[100010];

int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        int minn=INF;
        for(int i=-1;i<=1;i++)
        {
            for(int j=-1;j<=1;j++)
            {
                int flag=1,cnt=abs(i)+abs(j);
                b[1]=a[1]+i;
                b[2]=a[2]+j;
                int tmp=b[2]-b[1];
                for(int k=3;k<=n;k++)
                {
                    b[k]=b[k-1]+tmp;
                    //cout<<abs(b[k]-a[k])<<' ';
                    if(abs(b[k]-a[k])>1)
                    {

                        flag=0;
                        break;
                    }
                    if(b[k]!=a[k]) cnt++;
                }
                if(flag)
                {
                    /*for(int k=1;k<=n;k++)
                        cout<<b[k]<<' ';
                    */
                    minn=min(minn,cnt);
                }

            }

        }
        if(minn!=INF)
            cout<<minn<<endl;
        else
            cout<<"-1"<<endl;
    }
    return 0;
}