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>
int main() {
  int n;
  std::cin >> n;
  std::vector<long long> imos0(n + 1);
  std::vector<long long> imos1(n + 1);
  auto add = [&](int l, int r, long long a, long long b) {
    imos0[l] += a;
    imos0[r] -= a;
    imos0[l] -= b * l;
    imos0[r] += b * l;
    imos1[l] += b;
    imos1[r] -= b;
  };
  std::vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (i <= a[i]) {
      add(0, a[i] - i, std::abs(a[i] - i), -1);
      add(a[i] - i, n - i, 0, 1);
      add(n - i, n, std::abs(a[i] - 0), -1);
    } else {
      add(0, n - i, std::abs(a[i] - i), 1);
      add(n - i, n - i + a[i], std::abs(a[i] - 0), -1);
      add(n - i + a[i], n, 0, 1);
    }
  }
  for (int i = 0; i < n; i++) {
    imos0[i + 1] += imos0[i];
    imos1[i + 1] += imos1[i];
  }
  std::pair<long long, int> ans(1e18, 1e9);
  for (int i = 0; i < n; i++) {
    long long cnt = imos0[i] + imos1[i] * i;
    ans = std::min(ans, std::make_pair(cnt, i));
  }
  std::cout << ans.first << " " << ans.second << std::endl;
}
