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 <image>.

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 p1, p2, ... pn, 
  * k = 1: shift pn, p1, ... pn - 1, 
  * ..., 
  * k = n - 1: shift p2, p3, ... pn, p1. 

Input

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

The second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ 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.
if 1==1:
    n=input()
    A=list(map(int,raw_input().strip().split(' ')))

    #for i in range(len(A)):
    #    A[i]-=1

    ans=0
    pos=0
    neg=0
    change=[0 for i in range(len(A))]

    for i in range(len(A)):
        ans+=abs(A[i]-i-1)
        if A[i]-1-i>0:
            pos+=1
        else:
            neg+=1
        if A[i]-i-1>0:
            change[i]=A[i]-i-1
        elif A[i]==i+1:
            change[i]=0
        else:
            if A[i]!=1:
                change[i]=A[i]+n-i-1
            else:
                change[i]=0
    MIN=ans
    index=0
    #print(ans)
    collect=[0 for i in range(n)]
    for x in range(len(change)):
        collect[change[x]]+=1
    #print(collect)

    #print(ans,pos,neg)    
    for s in range(1,n):  
        ans-=abs(A[n-s]-n+1-1)
        ans+=abs(A[n-s]-0-1)
        neg-=1

        ans-=pos
        ans+=neg
        if A[n-s]>1:
            pos+=1
        else:
            neg+=1


        pos-=(collect[s])
        neg+=(collect[s])
        #print(ans,pos,neg)
        if ans<MIN:
            MIN=ans
            index=s
    print MIN,index      
    #brutal(A)        
#main()

    