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;
constexpr int N = 2 * ((int)3e6 + 10);
long long p[N], n;
int d[N + N];
long long cnt_left, cnt_right;
long long add_left, add_right;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[i];
  int mid = N / 2;
  long long cur_ans = 0;
  int step = 0;
  for (int i = 1; i <= n; ++i) {
    d[mid + p[i] - i]++;
    if (p[i] < i)
      cnt_left++, add_left += i - p[i];
    else if (p[i] > i)
      cnt_right++, add_right += p[i] - i;
    cur_ans += abs(p[i] - i);
  }
  for (int i = 1; i <= n; ++i) {
    d[mid + p[i] - 1]--;
    if (p[i] > 1) {
      cnt_right--;
      add_right -= p[i] - 1;
    } else if (p[i] < 1) {
      cnt_left--;
      add_left -= 1 - p[i];
    }
    cnt_right += d[mid];
    add_right += cnt_right;
    add_left -= cnt_left;
    cnt_left -= d[mid - 1];
    mid--;
    d[mid + p[i] - n]++;
    if (p[i] < n)
      cnt_left++, add_left += n - p[i];
    else if (p[i] > n)
      cnt_right++, add_right += p[i] - n;
    if (cur_ans > add_left + add_right) {
      cur_ans = add_left + add_right;
      step = n - i;
    }
  }
  cout << cur_ans << " " << step << endl;
  return 0;
}
