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 int arr[1000005];
long long int chg[5000005];
int cmpfunc(const void *a, const void *b) {
  return (*(const long long int *)a) - (*(const long long int *)b);
}
int main(void) {
  long long int i, j, k, n, ans, min, mini, grad, pt;
  scanf("%lli", &n);
  min = 0;
  grad = 0;
  for (i = 1; i <= n; i++) {
    scanf("%lli", &arr[i]);
    min += arr[i] > i ? arr[i] - i : i - arr[i];
  }
  for (i = 1; i < n; i++) {
    if (arr[i] > i) {
      grad--;
    } else {
      grad++;
    }
  }
  mini = 0;
  ans = min;
  for (i = 1; i <= n; i++) {
    chg[i - 1] = ((arr[i] - i) + n - 1) % n + 1;
  }
  qsort(chg, n, sizeof(chg[0]), cmpfunc);
  pt = 0;
  for (i = 1; i < n; i++) {
    min += grad;
    min += (2 * arr[n - i + 1] - n - 1);
    grad -= 2;
    while (chg[pt] <= i && pt < n) {
      pt++;
      grad += 2;
    }
    if (min < ans) {
      ans = min;
      mini = i;
    }
  }
  printf("%lli %lli\n", ans, mini);
}
