CF819B Mister B and PR Shifts

题目

Some time ago Mister B detected a strange signal from the space, which he started to study.

After some transformation the signal turned out to be a permutation p of length n or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.

Let's define the deviation of a permutation p as .

Find a cyclic shift of permutation p with minimum possible deviation. If there are multiple solutions, print any of them.

Let's denote id k (0 ≤ k < n) of a cyclic shift of permutation p as the number of right shifts needed to reach this shift, for example:

k = 0: shift p 1, p 2, ... p n, k = 1: shift p n, p 1, ... p n - 1, ..., k = n - 1: shift p 2, p 3, ... p n, p 1.

Input

First line contains single integer n (2 ≤ n ≤ 106) — the length of the permutation.

The second line contains n space-separated integers p 1, p 2, ..., p n (1 ≤ p i ≤ n) — the elements of the permutation. It is guaranteed that all elements are distinct.

Output

Print two integers: the minimum deviation of cyclic shifts of permutation p and the id of such shift. If there are multiple solutions, print any of them.

Examples

Input

3
1 2 3

Output

0 0

Input

3
2 3 1

Output

0 1

Input

3
3 2 1

Output

2 1

Note

In the first sample test the given permutation p is the identity permutation, that's why its deviation equals to 0, the shift id equals to 0 as well.

In the second sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 2, 3) equals to 0, the deviation of the 2-nd cyclic shift (3, 1, 2) equals to 4, the optimal is the 1-st cyclic shift.

In the third sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 3, 2) equals to 2, the deviation of the 2-nd cyclic shift (2, 1, 3) also equals to 2, so the optimal are both 1-st and 2-nd cyclic shifts.

题意

定义一个全排列 pi 的偏移值为∑ni=1| pi - i | 给你一个全排列，你可以从后面拿k∈[0,n−1]个数放在前面，使得该全排列的偏移值最小，输出这个偏移值和k，如果有多个k任意输出一个

分析

我们当前位的数的本身的大小是不变的，而我们每操作一次它的编号 i 就会增大1 。

那么，对于这次操作偏移值为正数的，结果的总贡献就回少1，也就是ans一共会减少原偏移值为正的数的个数。

对于此次操作偏移值为非正的，显然对ans的贡献会加1，所以ans的增加量是原偏移值为非正数的数的个数。

所以，我们要找的就是原偏移值为正的和非正的数的个数。

然后我们就可以考虑了，经过一次操作，原偏移值为+1的数偏移值都会变为0，所以，原偏移值为正数的数的个数会少掉原偏移值为+1的数的个数。

而原偏移值为非正数的数的个数会加上原偏移值为+1的数的个数，这也是显然的。所以原偏移值为非正的数的个数不会变少，只会越来越多。

最后每次我们把第 n-i+1 位的数放到最前面，然后他的改变就直接算就行。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
const int maxn=3e6+10;
int a[maxn];
ll ans,k;
ll cnt,sum;
ll one[maxn];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        int x=a[i]-i;
        if(x<=0){
            sum++;
            ans+=abs(x);
        }
        else{
            cnt++;
            ans+=x;
            one[x]++;
        }
    }
    ll tmp,tmpp=ans;
    if(ans==0){
        printf("0 0
");
        return 0;
    }
    for(int i=1;i<n;i++){
        tmp=tmpp;
        tmp-=cnt;
        tmp+=sum;
        cnt-=one[i];
        sum+=one[i];
        ll x=a[n-i+1];
        tmp-=n+1-x;
        sum--;
        if(x>1){
            one[x-1+i]++;
            tmp+=x-1;
            cnt++;
        }
        else sum++;
        if(tmp<ans){
            ans=tmp;
            k=i;
        }
        if(ans==0){
            printf("%d %d
",ans,k);
            return 0;
        }
        tmpp=tmp;
    }
    printf("%d %d
",ans,k);
    return 0;
}