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() {
  long long n;
  std::cin >> n;
  long long inversion = 0;
  std::vector<long long> a;
  a.reserve(n);
  std::vector<long long> chan(n, 0);
  long long total = 0;
  for (long long i = 0; i < n; ++i) {
    long long b;
    std::cin >> b;
    a.push_back(b);
    inversion += abs(b - i - 1);
    if (b > i + 1) {
      total++;
    }
    ++chan[(b - i - 1 + n) % n];
  }
  long long min = inversion;
  long long id = 0;
  for (long long i = n - 1; i > 0; --i) {
    inversion -= total;
    inversion += n - total;
    inversion -= n + 1 - a[i];
    inversion += a[i] - 1;
    total -= chan[n - i] - 1;
    if (min > inversion) {
      min = inversion;
      id = n - i;
    }
  }
  std::cout << min << " " << id << std::endl;
}
