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.

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).

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).

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.

枚举 a[1]、a[2]的所有情况，一但a[1]-a[2]确定，则所有数之间的差确定，看修改元素是否能达到这种可能。如果可以，维护一个最小值。

#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
#define maxn 100005
using namespace std;
typedef long long ll;
int main()
{int n,i,j,k;int a[maxn]={0};int b[maxn]={0};scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}if(n==2||n==1){printf("0\n");return 0;}int minum=10000000;int flag=0;for(i=-1;i<=1;i++){for(j=-1;j<=1;j++){int cnt=abs(i)+abs(j);for(k=1;k<=n;k++){b[k]=a[k];}b[1]+=i;b[2]+=j;int x=b[1]-b[2];for(k=2;k<=n-1;k++){if(b[k]-(b[k+1]+1)==x){b[k+1]+=1;cnt++;}else if(b[k]-b[k+1]==x){}else if(b[k]-(b[k+1]-1)==x){cnt++;b[k+1]-=1;}else{break;}if(k==n-1){minum=min(minum,cnt);flag=1;}}}}if(flag) printf("%d\n",minum);else printf("-1\n");return 0;
}