#include <bits/stdc++.h>
using namespace std;
int nbr[1001000], movx[1001000];
int n;
int main() {
  int i;
  while (scanf("%d", &n) > 0) {
    memset(movx, 0, sizeof(movx));
    long long int ans = 0, resn = 0;
    int left = 0, right = 0;
    int resp = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &nbr[i]);
      resn += abs(nbr[i] - i);
      if (nbr[i] >= i)
        movx[nbr[i] - i]++;
      else
        movx[n - i + nbr[i]]++;
      if (nbr[i] > i)
        left++;
      else
        right++;
    }
    ans = resn;
    for (i = 1; i < n; i++) {
      ans += right;
      ans -= left;
      ans += (nbr[n - i + 1] - 1) - (n + 1 - nbr[n - i + 1]);
      left -= movx[i];
      right += movx[i];
      left++;
      right--;
      if (ans < resn) {
        resn = ans;
        resp = i;
      }
    }
    printf("%lld %d\n", resn, resp);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
using namespace std;
int nbr[1001000], movx[1001000];
int n;
int main() {
  int i;
  while (scanf("%d", &n) > 0) {
    memset(movx, 0, sizeof(movx));
    long long int ans = 0, resn = 0;
    int left = 0, right = 0;
    int resp = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &nbr[i]);
      resn += abs(nbr[i] - i);
      if (nbr[i] >= i)
        movx[nbr[i] - i]++;
      else
        movx[n - i + nbr[i]]++;
      if (nbr[i] > i)
        left++;
      else
        right++;
    }
    ans = resn;
    for (i = 1; i < n; i++) {
      ans += right;
      ans -= left;
      ans += (nbr[n - i + 1] - 1) - (n + 1 - nbr[n - i + 1]);
      left -= movx[i];
      right += movx[i];
      left++;
      right--;
      if (ans < resn) {
        resn = ans;
        resp = i;
      }
    }
    printf("%lld %d\n", resn, resp);
  }
  return 0;
}

```