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>
using namespace std;
long long ans[1000005], num[1000005];
int main() {
  int n;
  long long minn, x;
  int temp;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    num[i] = x;
    if (i < x) {
      ans[1]--;
      ans[x - i + 1]++;
      ans[x - i + 1]++;
      ans[n - i + 1]--;
      ans[n - i + 1] += 2 * x - 1 - n;
      ans[n - i + 2] -= 2 * x - 1 - n;
      ans[n - i + 2]--;
    } else {
      ans[1]++;
      ans[n - i + 1]--;
      ans[n - i + 1] += 2 * x - 1 - n;
      ans[n - i + 2] -= 2 * x - 1 - n;
      ans[n - i + 2]--;
      ans[n - i + x + 1]++;
      ans[n - i + x + 1]++;
    }
  }
  for (int i = 1; i < n; i++) ans[i] += ans[i - 1];
  for (int i = 1; i <= n; i++) ans[0] += abs(i - num[i]);
  temp = 0;
  minn = ans[0];
  for (int i = 1; i < n; i++) {
    ans[i] += ans[i - 1];
    if (ans[i] < minn) {
      minn = ans[i];
      temp = i;
    }
  }
  printf("%lld %d\n", minn, temp);
  return 0;
}
