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>
const int maxn = 1e6 + 1e2;
int val[maxn];
int derivative1[maxn];
int derivative2[maxn];
long long curr = 0;
long long curr_change = 0;
long long res = 1e18;
long long res_pos = 0;
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  std::cin >> n;
  for (int i = 0; i < n; i++) {
    std::cin >> val[i];
    val[i]--;
    curr += std::abs(val[i] - i);
    if (i <= val[i]) {
      derivative2[1]--;
      derivative2[val[i] - i + 1] += 2;
      derivative2[n - i]--;
      derivative1[n - i] += std::abs(val[i] - 0) - std::abs(val[i] - (n - 1));
      derivative2[n - i + 1]--;
    } else {
      derivative2[1]++;
      derivative2[n - i]--;
      derivative1[n - i] += std::abs(val[i] - 0) - std::abs(val[i] - (n - 1));
      derivative2[n - i + 1]--;
      derivative2[n - i + val[i] + 1] += 2;
    }
  }
  for (int i = 0; i < n; i++) {
    curr_change += derivative2[i];
    curr += curr_change + derivative1[i];
    if (res > curr) {
      res = curr;
      res_pos = i;
    }
  }
  std::cout << res << ' ' << res_pos << std::endl;
}
