#include <bits/stdc++.h>
signed main() {
  long long n;
  std::cin >> n;
  std::vector<long long> seq(n);
  for (long long i = 0; i < n; i++) {
    std::cin >> seq[i];
    seq[i]--;
  }
  std::vector<long long> p(2 * n, 0);
  std::vector<long long> q(2 * n, 0);
  long long l, r;
  for (long long i = 0; i < n; i++) {
    l = i - seq[i];
    r = i;
    if (l < 0) {
      l += n;
      r += n;
    }
    p[l] += seq[i] - r;
    if (r + 1 < 2 * n) p[r + 1] -= seq[i] - r;
    q[l]++;
    if (r + 1 < 2 * n) q[r + 1]--;
    if (seq[i] == n - 1) continue;
    l = i + 1;
    r = i - seq[i] - 1 + n;
    p[l] -= seq[i] - l - n + 1;
    if (r + 1 < 2 * n) p[r + 1] += seq[i] - l - n + 1;
    q[l]--;
    if (r + 1 < 2 * n) q[r + 1]++;
  }
  std::vector<long long> ans(n, 0);
  for (long long i = 0; i < 2 * n; i++) {
    if (i) p[i] += p[i - 1];
    if (i) q[i] += q[i - 1];
    ans[i % n] += p[i] + q[i] * i;
  }
  long long min, id;
  min = ans[0];
  id = 0;
  for (long long i = 1; i < n; i++) {
    if (ans[i] < min) {
      min = ans[i];
      id = i;
    }
  }
  std::cout << min << " " << (n - id) % n << std::endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
signed main() {
  long long n;
  std::cin >> n;
  std::vector<long long> seq(n);
  for (long long i = 0; i < n; i++) {
    std::cin >> seq[i];
    seq[i]--;
  }
  std::vector<long long> p(2 * n, 0);
  std::vector<long long> q(2 * n, 0);
  long long l, r;
  for (long long i = 0; i < n; i++) {
    l = i - seq[i];
    r = i;
    if (l < 0) {
      l += n;
      r += n;
    }
    p[l] += seq[i] - r;
    if (r + 1 < 2 * n) p[r + 1] -= seq[i] - r;
    q[l]++;
    if (r + 1 < 2 * n) q[r + 1]--;
    if (seq[i] == n - 1) continue;
    l = i + 1;
    r = i - seq[i] - 1 + n;
    p[l] -= seq[i] - l - n + 1;
    if (r + 1 < 2 * n) p[r + 1] += seq[i] - l - n + 1;
    q[l]--;
    if (r + 1 < 2 * n) q[r + 1]++;
  }
  std::vector<long long> ans(n, 0);
  for (long long i = 0; i < 2 * n; i++) {
    if (i) p[i] += p[i - 1];
    if (i) q[i] += q[i - 1];
    ans[i % n] += p[i] + q[i] * i;
  }
  long long min, id;
  min = ans[0];
  id = 0;
  for (long long i = 1; i < n; i++) {
    if (ans[i] < min) {
      min = ans[i];
      id = i;
    }
  }
  std::cout << min << " " << (n - id) % n << std::endl;
  return 0;
}

```