CF819B.Mister B and PR Shifts

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题目描述

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 pp of length nn 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 pp as .

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

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

  • k=0k=0 : shift p1,p2,... pnp_{1},p_{2},...\ p_{n} ,
  • k=1k=1 : shift pn,p1,... pn1p_{n},p_{1},...\ p_{n-1} ,
  • ...,
  • k=n1k=n-1 : shift p2,p3,... pn,p1p_{2},p_{3},...\ p_{n},p_{1} .

输入格式

First line contains single integer nn ( 2<=n<=1062<=n<=10^{6} ) — the length of the permutation.

The second line contains nn space-separated integers p1,p2,...,pnp_{1},p_{2},...,p_{n} ( 1<=pi<=n1<=p_{i}<=n ) — the elements of the permutation. It is guaranteed that all elements are distinct.

输出格式

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

输入输出样例

  • 输入#1

    3
    1 2 3
    

    输出#1

    0 0
    
  • 输入#2

    3
    2 3 1
    

    输出#2

    0 1
    
  • 输入#3

    3
    3 2 1
    

    输出#3

    2 1
    

说明/提示

In the first sample test the given permutation pp is the identity permutation, that's why its deviation equals to 00 , the shift id equals to 00 as well.

In the second sample test the deviation of pp equals to 44 , the deviation of the 11 -st cyclic shift (1,2,3)(1,2,3) equals to 00 , the deviation of the 22 -nd cyclic shift (3,1,2)(3,1,2) equals to 44 , the optimal is the 11 -st cyclic shift.

In the third sample test the deviation of pp equals to 44 , the deviation of the 11 -st cyclic shift (1,3,2)(1,3,2) equals to 22 , the deviation of the 22 -nd cyclic shift (2,1,3)(2,1,3) also equals to 22 , so the optimal are both 11 -st and 22 -nd cyclic shifts.

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