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;
template <class T>
struct BinaryIndexedTree {
  vector<T> data;
  BinaryIndexedTree(int sz) { data.assign(++sz, 0); }
  T sum(int k) {
    T ret = 0;
    for (++k; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }
  void add(int k, T x) {
    for (++k; k < data.size(); k += k & -k) data[k] += x;
  }
};
int main() {
  int N, P[1000000];
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &P[i]), --P[i];
  int T[1000000];
  for (int i = 0; i < N; i++) T[i] = P[i] - i;
  long long ret = numeric_limits<long long>::max();
  int id;
  long long now = 0LL;
  for (int i = 0; i < N; i++) now += abs(T[i]);
  BinaryIndexedTree<int> bit(N + N + N + 1);
  for (int i = 0; i < N; i++) T[i] += N;
  for (int i = 0; i < N; i++) bit.add(T[i], 1);
  for (int i = 0; i < N; i++) {
    if (ret > now) {
      ret = now;
      id = i;
    }
    int base_line = i + N + 1;
    now += bit.sum(base_line - 1);
    now -= bit.sum(N + N + N) - bit.sum(base_line - 1);
    int get = T[N - i - 1];
    bit.add(get, -1);
    now -= abs(get - base_line);
    bit.add(get + N, 1);
    now += abs(N + get - base_line);
  }
  cout << ret << " " << id << endl;
}
