820_D. Mister B and PR Shifts
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.
```python
from sys import stdin
def main():
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().split()))
    inf = [0] * (n + 1)
    curr = 0
    d = 0
    for i in range(n):
        curr += abs(i + 1 - a[i])
        if a[i] > i + 1:
            d += 1
            inf[a[i] - i - 1] += 1
        elif a[i] <= i + 1:
            d -= 1
            if a[i] == i + 1:
                inf[0] += 1
            else:
                inf[a[i] + n - i - 1] += 1
    best = curr
    num = 0
    for i in range(n):
        curr -= d
        curr -= 1
        curr = curr - abs(a[n - i - 1] - n) + abs(a[n - i - 1] - 1)
        d += 2
        d -= inf[i + 1] * 2
        if curr < best:
            best = curr
            num = i + 1
    print(best, num)
    
main()
```