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;
using ll = long long;
using ld = long double;
using ii = pair<int, int>;
using vi = vector<int>;
constexpr int N = 1000010;
constexpr ll INFL = 1000000000000000000;
int n, p[N];
int a[2 * N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(15);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[i];
  int pos = 0;
  ll ans = 0, ansj = 0;
  for (int i = 1; i <= n; ++i) {
    ans += abs(p[i] - i);
    if (p[i] - i > 0) {
      ++pos;
      ++a[p[i] - i];
    }
  }
  ll cur = ans;
  for (int j = 1; j < n; ++j) {
    int ind = n + 1 - j;
    cur += n - 1 - 2 * pos + (abs(p[ind] - 1) - abs(p[ind] - n));
    if (cur < ans) {
      ans = cur;
      ansj = j;
    }
    if (p[ind] - ind > 0) --a[p[ind] - ind];
    pos -= a[j];
    if (p[ind] - 1 > 0) {
      ++pos;
      ++a[p[ind] - 1 + j];
    }
  }
  cout << ans << ' ' << ansj;
}
