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;
long long A[1000010], B[1000010], C[1000010], final[1000010], n;
void addap(int left, int right, int dif, int start) {
  if (left > right) return;
  C[left] += dif;
  B[left] += start;
  C[right] -= dif;
  B[right + 1] -= start + 1LL * dif * (right - left);
}
void prefix() {
  long long cur = 0;
  for (int i = 0; i < n; i++) {
    B[i] += cur;
    cur += C[i];
  }
  long long sm = 0;
  for (int i = 0; i < n; i++) {
    sm += B[i];
    final[i] += sm;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= n; i++) {
    if (A[i] > i) {
      addap(0, A[i] - i, -1, A[i] - i);
      addap(A[i] - i + 1, n - i, 1, 1);
      addap(n - i + 1, n - 1, -1, A[i] - 1);
    } else {
      addap(0, n - i, 1, i - A[i]);
      addap(n - i + 1, n - i + A[i], -1, A[i] - 1);
      addap(n - i + A[i] + 1, n - 1, 1, 1);
    }
  }
  prefix();
  long long mini = 1000000000, ind = -1;
  for (int i = 0; i < n; i++) {
    if (final[i] < mini || ind == -1) {
      mini = final[i];
      ind = i;
    }
  }
  cout << mini << " " << ind << '\n';
}
