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.
#include <bits/stdc++.h>
long long minimal_shift(int* array, int len) {
  int diviation_len = len * 2 - 1;
  int* diviation = (int*)calloc(diviation_len, sizeof(diviation[0]));
  int middle = diviation_len / 2;
  int up = 0;
  long long diff = 0;
  for (int i = 0; i < len; i++) {
    int index = array[i] - i - 1;
    if (index > 0) up++;
    diff += abs(index);
    diviation[middle + index] += 1;
  }
  int down = up;
  int equal = len - up - down;
  long long min_diff = diff;
  long long shift = 0;
  for (int i = 0; i < len - 1; i++) {
    diff -= up;
    diff += equal;
    diff += down;
    diff -= abs(array[len - i - 1] - len);
    diff += array[len - i - 1] - 1;
    int tmp = array[len - i - 1] - 1;
    if (middle + i + tmp + 1 < diviation_len) {
      diviation[middle + i + tmp + 1]++;
    }
    down += equal;
    equal = diviation[middle + i + 1];
    up -= equal;
    if (array[len - i - 1] == len) {
      down--;
      diff--;
      up++;
    } else {
      diff--;
      down--;
      up++;
    }
    if (min_diff > diff) {
      min_diff = diff;
      shift = i + 1;
    }
  }
  free(diviation);
  return shift;
}
int main(void) {
  int len;
  std::cin >> len;
  int* array = (int*)malloc(sizeof(array[0]) * len);
  for (int i = 0; i < len; i++) std::cin >> array[i];
  long long shift = minimal_shift(array, len);
  long long diff = 0;
  for (int i = 0; i < len; i++) diff += abs(array[i] - (i + shift) % len - 1);
  std::cout << diff << " " << shift << std::endl;
  free(array);
  return 0;
}
