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.
n = int(raw_input())
P = map(lambda x: int(x)-1, raw_input().split())

def deviation(P):
  res = 0
  for i in range(len(P)):
    res += abs(i - P[i])
  return res

def shift(P, n):
  return P[len(P) - n:] + P[:len(P) - n]

def gold(P, n):
  best, best_id = deviation(P), 0
  for k in range(1, n):
    tmp = deviation(shift(P, k))
    if tmp < best:
      best, best_id = tmp, k
  return best, best_id


d = 0
neg = 0
zer = 0
pos = 0

from collections import defaultdict

C = [[] for _ in range(n)]

for i in range(n):
  d += abs(i - P[i])
  if i == P[i]: zer += 1
  if i < P[i]: neg += 1
  if i > P[i]: pos += 1
  delta = (P[i] - i)%n
  C[delta].append(i)

best, best_id = d, 0
for k in range(1, n):
  # First manage the case of the rightmost one: move it at pos "-1"
  if P[n-k] == n-1: zer -= 1
  else: pos -= 1
  neg += 1
  d -= (n-1) - P[n-k]
  d += P[n-k] + 1

  # Every zeroed is becoming positive
  pos += zer
  zer = 0
  # Every positive increments its deviation
  d += pos
  # Every negative decrements its deviation
  d -= neg

  # Some negatives are becoming zero
  neg -= len(C[k])
  zer += len(C[k])

  if d < best:
    best, best_id = d, k
print best, best_id
