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;
const long double pi = 3.141592653589793;
int n;
long long P[1 << 20];
long long H[1 << 20];
void update(int i, int j, int inr) {
  H[i] += inr;
  H[j + 1] -= inr;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  long long val = 0, which = 0;
  for (int i = 0; i < n; i++) cin >> P[i + 1];
  for (int k = 1; k <= n; k++) {
    val += abs(P[k] - k);
    if (P[k] >= k) {
      int s = 1, diff;
      diff = P[k] - k;
      if (diff > 0) {
        update(s, s + diff - 1, -1);
        s = s + diff;
      }
      if (s == n) continue;
      diff = (n - P[k]);
      if (diff > 0) {
        update(s, s + diff - 1, 1);
        s = s + diff;
      }
      if (s == n) continue;
      diff = 1;
      if (n - 2 * P[k] + 1 > 0)
        update(s, s + diff - 1, -(n - 2 * P[k] + 1));
      else
        update(s, s + diff - 1, -(n - 2 * P[k] + 1));
      s = s + diff;
      if (s == n) continue;
      if (k - 2 > 0) update(s, n - 1, -1);
    } else {
      int s = 1, diff;
      diff = n - k;
      if (diff > 0) {
        update(s, s + diff - 1, 1);
        s = s + diff;
      }
      if (s == n) continue;
      diff = 1;
      if (n - 2 * P[k] + 1 > 0)
        update(s, s + diff - 1, -(n - 2 * P[k] + 1));
      else
        update(s, s + diff - 1, -(n - 2 * P[k] + 1));
      s = s + diff;
      if (s == n) continue;
      diff = (P[k] - 1);
      if (diff > 0) {
        update(s, s + diff - 1, -1);
        s = s + diff;
      }
      if (s == n) continue;
      if (k - 1 - P[k] > 0) update(s, n - 1, 1);
    }
  }
  for (int i = 1; i < n; i++) H[i] += H[i - 1];
  long long ans = val;
  for (int i = 1; i < n; i++) {
    val = val + H[i];
    if (val < ans) {
      ans = val;
      which = i;
    }
  }
  cout << ans << " " << which << endl;
  return 0;
}
