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 int N = 1e6 + 5;
const long long INF = 1e18;
int n;
int p[N];
long long dif[N];
long long ans = INF;
int id;
void add(int l, int r, int val) {
  dif[l] += val;
  dif[r + 1] -= val;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    res += (1LL) * abs(p[i] - i);
    if (p[i] >= i) {
      add(2, i, 1);
      add(i + 1, i + 1, (n - p[i]) - (p[i] - 1));
      add(i + 2, i + 2 + (n - p[i]) - 1, -1);
      add(i + 2 + (n - p[i]), n, 1);
    } else {
      add(2, i - p[i] + 1, -1);
      add(i - p[i] + 2, i, 1);
      add(i + 1, i + 1, (n - p[i]) - (p[i] - 1));
      add(i + 2, n, -1);
    }
  }
  for (int i = 1; i <= n; i++) {
    dif[i] += dif[i - 1];
    res += dif[i];
    if (res < ans) {
      ans = res, id = n - i + 1;
      if (id == n) id = 0;
    }
  }
  cout << ans << " " << id << '\n';
  return 0;
}
