#include <bits/stdc++.h>
using namespace std;
int n;
int a[1000000];
int v[1000001];
int main() {
  long long nowsum = 0;
  long long offset = 0;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    nowsum += abs(a[i] - (i + 1));
    if (a[i] < i + 1) {
      v[i + 1 - a[i]]++;
      offset--;
    } else {
      offset++;
    }
  }
  ans = nowsum;
  int r = 0;
  for (int i = 0; i < n - 1; i++) {
    nowsum += offset;
    nowsum -= a[i];
    nowsum += abs(a[i] - n);
    if (a[i] != n) {
      offset -= 2;
      if (n - a[i] + i < n) v[n - a[i] + i + 1]++;
    }
    offset += 2 * v[i + 1];
    if (ans > nowsum) {
      ans = nowsum;
      r = n - 1 - i;
    }
  }
  printf("%I64d %d\n", ans, r);
  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>
using namespace std;
int n;
int a[1000000];
int v[1000001];
int main() {
  long long nowsum = 0;
  long long offset = 0;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    nowsum += abs(a[i] - (i + 1));
    if (a[i] < i + 1) {
      v[i + 1 - a[i]]++;
      offset--;
    } else {
      offset++;
    }
  }
  ans = nowsum;
  int r = 0;
  for (int i = 0; i < n - 1; i++) {
    nowsum += offset;
    nowsum -= a[i];
    nowsum += abs(a[i] - n);
    if (a[i] != n) {
      offset -= 2;
      if (n - a[i] + i < n) v[n - a[i] + i + 1]++;
    }
    offset += 2 * v[i + 1];
    if (ans > nowsum) {
      ans = nowsum;
      r = n - 1 - i;
    }
  }
  printf("%I64d %d\n", ans, r);
  return 0;
}

```