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;
int A[1000001];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  map<int, int> change_v;
  int above = 0, below = 0;
  long long sum = 0;
  long long best_i = 0;
  for (int i = 0; i < n; i++) {
    int xs = A[i] - i - 1;
    int x = (xs > 0) ? xs : -1 * xs;
    if (xs < 0) {
      change_v[x]++;
    }
    if (xs < 0) {
      below++;
    } else {
      above++;
    }
    sum += x;
  }
  long long best_sum = sum;
  for (int i = 0; i < n - 1; i++) {
    long long xs = A[i] - n;
    long long x = (xs > 0) ? xs : -1 * xs;
    long long cnt = change_v[i + 1];
    if (xs < 0) {
      change_v[x + i + 1]++;
    }
    int cur_below_d = (A[i] - 1 < 0) ? 1 : 0;
    int cur_above_d = (A[i] - 1 >= 0) ? 1 : 0;
    sum -= below - cur_below_d;
    sum += above - cur_above_d;
    sum -= (A[i] - 1 > 0) ? A[i] - 1 : 1 - A[i];
    sum += x;
    above += cnt;
    below -= cnt;
    if (xs < 0 && A[i] - 1 >= 0) {
      below++;
      above--;
    }
    if (sum < best_sum) {
      best_i = i + 1;
      best_sum = sum;
    }
  }
  if (best_i != 0) {
    best_i = n - best_i;
  }
  cout << best_sum << " " << best_i;
}
