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 T[1000000 * 3];
int K[1000000 * 3];
int ad(int k) {
  if (k > 0)
    return k;
  else
    return -k;
}
int main() {
  int n;
  scanf("%d", &n);
  int posi = 0;
  int neg = 0;
  long long sum = 0;
  int d;
  memset(T, 0, sizeof(T));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d);
    K[i] = d;
    if (d - i > 0) {
      posi++;
      T[d - i]++;
    } else if (d - i <= 0)
      neg++;
    sum = sum + ad(d - i);
  }
  long long mins = sum;
  int kp = 0;
  for (int k = 1; k < n; k++) {
    sum = sum - abs(K[n - k + 1] - n) + abs(K[n - k + 1] - 1);
    neg--;
    neg = max(0, neg);
    sum = sum - posi + neg;
    T[K[n - k + 1] - (n - k + 1)]--;
    if (K[n - k + 1] != 1) {
      T[K[n - k + 1] - 1 + k]++;
      posi++;
    } else
      neg++;
    posi = posi - T[k];
    neg = neg + T[k];
    if (mins > sum) {
      kp = k;
      mins = sum;
    }
  }
  cout << mins << ' ' << kp << endl;
  return 0;
}
