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 N;
int arr[1000005];
priority_queue<int, vector<int>, greater<int>> pq;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  long long crnt = 0, add = 0, sub = 0;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
    crnt += abs(i - arr[i]);
    if (arr[i] > i) {
      add++;
    } else {
      sub++;
    }
    pq.push(i - arr[i] + 1 <= 0 ? i + N - arr[i] + 1 : i - arr[i] + 1);
  }
  long long ans = LLONG_MAX / 2, idx = 0;
  for (int i = 1; i <= N; i++) {
    if (ans > crnt) {
      idx = i - 1;
      ans = crnt;
    }
    while (pq.size() && pq.top() == i) {
      pq.pop();
      sub--;
      add++;
    }
    crnt -= arr[i] - 1;
    crnt += N - arr[i];
    sub++;
    add--;
    crnt += add - sub + 1;
  }
  cout << ans << " " << (N - idx) % N << endl;
}
