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 main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long p[n], to_zero[n], numl = 0, numg = 0;
  fill(to_zero, to_zero + n, 0);
  for (int i = 0; i < n; i++) cin >> p[i], p[i]--;
  for (int i = 0; i < n; i++) {
    if (i <= p[i])
      to_zero[p[i] - i]++;
    else if (i > p[i])
      to_zero[n - i + p[i]]++;
  }
  int shift = 0;
  long long curr = 0, min_ans;
  for (long long i = 0; i < n; i++) {
    curr += labs(p[i] - i);
    if (i < p[i])
      numl++;
    else
      numg++;
  }
  min_ans = curr;
  for (long long k = 1; k < n; k++) {
    curr += numg - numl + 2 * p[n - k] - n;
    numl -= to_zero[k] - 1;
    numg = n - numl;
    if (curr < min_ans) min_ans = curr, shift = k;
  }
  cout << min_ans << " " << shift << endl;
  ;
  return 0;
}
