1750.Mister B and PR Shifts

Time Limit: 1s Memory Limit: 256MB

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 1750_1.png.

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 \le k \lt n) of a cyclic shift of permutation p as the number of right shifts needed to reach this shift, for example:
k=0: shift p1,p2,... pn, k=1: shift pn,p1,... pn-1, ..., k=n-1: shift p2,p3,... pn,p1.

Input Format(From the terminal/stdin)

First line contains single integer n (2 \le n \le 106) - the length of the permutation.

The second line contains n space-separated integers p1,p2,...,pn (1 \le pi \le n)- the elements of the permutation. It is guaranteed that all elements are distinct.

Output Format(To the terminal/stdout)

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.

Sample Input 1

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3
1 2 3
 \n
 · · \n

Sample Output 1

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0 0
 · \n

Sample Input 2

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3
2 3 1
 \n
 · · \n

Sample Output 2

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0 1
 · \n

Sample Input 3

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3
3 2 1
 \n
 · · \n

Sample Output 3

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2 1
 · \n

Hints

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.

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