819_B. Mister B and PR Shifts_36788
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

3
1 2 3
 3
3 2 1
 3
2 3 1
 10
2 5 10 3 6 4 9 1 8 7
 108
1 102 33 99 6 83 4 20 61 100 76 71 44 9 24 87 57 2 81 82 90 85 12 30 66 53 47 36 43 29 31 64 96 84 77 23 93 78 58 68 42 55 13 70 62 19 92 14 10 65 63 75 91 48 11 105 37 50 32 94 18 26 52 89 104 106 86 97 80 95 17 72 40 22 79 103 25 101 35 51 15 98 67 5 34 69 54 27 45 88 56 16 46 60 74 108 21 41 73 39 107 59 3 8 28 49 7 38
 4
4 2 1 3
 10
10 1 9 2 8 3 7 4 6 5
 10
2 7 10 1 6 3 4 8 9 5
 10
1 2 3 4 6 5 7 9 10 8
 4
4 2 3 1
 4
3 1 4 2
 10
2 3 10 5 4 8 6 9 7 1
 10
1 2 10 9 7 4 8 3 6 5
 4
1 3 4 2
 4
3 1 2 4
 10
2 1 10 5 8 4 9 3 7 6
 4
3 4 1 2
 10
1 8 10 6 2 4 9 3 7 5
 4
4 1 3 2
 10
1 6 10 7 9 5 3 8 4 2
 10
2 4 10 3 9 1 5 7 8 6
 4
2 4 3 1
 4
2 3 1 4
 4
3 4 2 1
 10
1 7 10 6 5 2 3 8 9 4
 10
1 5 10 8 4 3 9 2 7 6
 4
3 2 4 1
 10
2 6 10 1 9 7 4 8 5 3
 4
2 3 4 1
 4
3 2 1 4
 2
1 2
 10
1 3 10 9 4 7 5 8 6 2
 4
4 3 1 2
 10
1 9 10 5 6 7 3 8 4 2
 2
2 1
 4
1 4 3 2
 4
2 1 4 3
 4
1 3 2 4
 4
4 1 2 3
 4
1 4 2 3
 10
1 10 9 5 3 2 4 7 8 6
 4
1 2 3 4
 10
1 4 10 8 9 2 3 6 7 5
 4
4 3 2 1
 4
2 1 3 4
 4
1 2 4 3
 4
2 4 1 3
 10
1 5 10 9 4 3 9 2 7 6
 4
3 2 4 2
 4
3 2 2 4
 4
1 2 4 1
 10
1 7 10 10 5 2 3 8 9 4
 4
1 3 3 4
 10
1 4 10 8 9 2 3 6 7 9
 4
4 3 1 3
 10
1 2 5 4 6 5 7 9 10 8
 10
2 4 10 6 9 1 5 7 8 6
 4
1 4 4 2
 4
1 4 2 4
 3
1 3 3

0 0
 2 1
 0 1
 28 0
 3428 30
 2 3
 24 7
 20 7
 6 0
 4 1
 4 1
 14 1
 26 5
 2 1
 2 3
 28 0
 0 2
 24 6
 2 3
 24 4
 28 0
 2 1
 4 0
 2 2
 26 6
 26 6
 2 1
 28 1
 0 1
 4 0
 0 0
 22 1
 2 2
 26 1
 0 1
 4 0
 4 0
 2 0
 0 3
 4 0
 20 7
 0 0
 20 5
 4 1
 2 0
 2 0
 2 2
 25 6
 3 1
 3 0
 2 1
 22 6
 1 0
 24 5
 3 2
 8 0
 25 5
 3 1
 3 0
 1 0
