codeforces

D. Almost Arithmetic Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

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

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

It is possible that the resulting sequence contains element equals $\mathrm{00\; .}$

Input

The first line contains a single integer $\mathrm{nn(1\le n\le 100000)(1\le n\le 100000)\; —\; the\; number\; of\; elements\; in\; bb\; .}$

The second line contains a sequence ${\mathrm{b1,b2,\dots ,bnb1,b2,\dots ,bn(1\le bi\le 109)(1\le bi\le 109)\; .}}^{}$

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

Copy

4
24 21 14 10

Output

Copy

3

Input

Copy

2
500 500

Output

Copy

0

Input

Copy

3
14 5 1

Output

Copy

-1

Input

Copy

5
1 3 6 9 12

Output

Copy

1

Note

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

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 $[0,3,6,9,12][0,3,6,9,12]\; ,\; which\; is\; an\; arithmetic\; progression.$

解题思路：等差数列的公差相等， 所以前两个数就可以确定公差，记录后面的数满足公差需要多少步即可，O(6 * n)的复杂度。

附ac代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <string>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <cstdlib>
const int maxn = 3 * 1e5 + 10;
const int inf = 0x3f3f3f3f;

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
int nu[maxn];
int tem[maxn];
queue<int>q;

int main(int argc, const char * argv[]) {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
    {
        scanf("%d", &nu[i]);
    }
    if(n <= 2)
    {
        puts("0");
        return 0;
    }
    int i, j, k;
    int ans = inf;
    for(i = -1; i <= 1; ++i)
    {
        for(j = -1; j <= 1; ++j)
        {
            tem[0] = nu[0] + i;
            tem[1] = nu[1] + j;
            int d = tem[1] - tem[0];
            int cnt = abs(i) + abs(j);
            for(k = 2; k < n; ++k)
            {
                int u = nu[k] - tem[k - 1];
                tem[k] = d + tem[k - 1];
                if(u == d) continue;
                if(abs(u - d) <= 1)
                {
                    cnt++;
                  //  printf("%d ", tem[k]);
                }
                else break;
            }
            if(k == n)
            {
                ans = min(cnt, ans);
            }
        }
    }
    if(ans == inf) puts("-1");
    else
    printf("%d
", ans);
    return 0;
}