#include <bits/stdc++.h>
long long p[1000007];
long long change[1000007];
long long what[1000007];
int main(int argc, char const *argv[]) {
  long long n;
  std::cin >> n;
  long long sum = 0;
  long long inc = 0;
  for (int i = 0; i <= n; ++i) {
    change[i] = 0;
    what[i] = -2;
  }
  for (int i = 1; i <= n; ++i) {
    std::cin >> p[i];
    sum += std::abs(p[i] - i);
    if (p[i] > i) {
      what[(p[i] - i + n) % n] += 2;
      inc -= 1;
    } else if (p[i] <= i) {
      what[(p[i] - i + n) % n] += 2;
      if (i != n) inc += 1;
    }
    change[n - i + 1] += (std::abs(p[i] - 1) - std::abs(p[i] - n));
  }
  int id = 0;
  long long temp = sum;
  for (int i = 1; i < n; ++i) {
    temp = temp + inc + change[i];
    if (sum > temp) {
      sum = temp;
      id = i;
    }
    inc += what[i];
  }
  std::cout << sum << " " << id << std::endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
long long p[1000007];
long long change[1000007];
long long what[1000007];
int main(int argc, char const *argv[]) {
  long long n;
  std::cin >> n;
  long long sum = 0;
  long long inc = 0;
  for (int i = 0; i <= n; ++i) {
    change[i] = 0;
    what[i] = -2;
  }
  for (int i = 1; i <= n; ++i) {
    std::cin >> p[i];
    sum += std::abs(p[i] - i);
    if (p[i] > i) {
      what[(p[i] - i + n) % n] += 2;
      inc -= 1;
    } else if (p[i] <= i) {
      what[(p[i] - i + n) % n] += 2;
      if (i != n) inc += 1;
    }
    change[n - i + 1] += (std::abs(p[i] - 1) - std::abs(p[i] - n));
  }
  int id = 0;
  long long temp = sum;
  for (int i = 1; i < n; ++i) {
    temp = temp + inc + change[i];
    if (sum > temp) {
      sum = temp;
      id = i;
    }
    inc += what[i];
  }
  std::cout << sum << " " << id << std::endl;
  return 0;
}

```