#include <bits/stdc++.h>
using namespace std;
int n;
int A[1000005];
int big[1000005];
int main() {
  int a = 0, b = 0, c = 0;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    ans += abs(A[i] - i);
    if (A[i] > i) {
      c++;
      big[A[i] - i]++;
    } else if (A[i] == i) {
      a++;
      big[0]++;
    } else {
      b++;
    }
  }
  int pos = n, id = 0;
  int ans_pos = 0;
  long long tmp = ans;
  while (--pos) {
    id++;
    tmp += b + a;
    tmp -= c;
    b = b + a;
    c = c - big[id];
    a = big[id];
    if (A[pos + 1] == n) {
      tmp += n - 2;
      int t = id + n - 1;
      if (t <= n) big[t]++;
      b--;
      c++;
    } else if (A[pos + 1] == 1) {
      b--;
      a++;
      tmp -= n;
    } else {
      b--;
      c++;
      int t = id + A[pos + 1] - 1;
      if (t <= n) big[t]++;
      tmp += -abs(A[pos + 1] - n - 1) + (A[pos + 1] - 1);
    }
    if (tmp <= ans) {
      ans = tmp;
      ans_pos = id;
    }
  }
  printf("%I64d %d\n", ans, ans_pos);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
int n;
int A[1000005];
int big[1000005];
int main() {
  int a = 0, b = 0, c = 0;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    ans += abs(A[i] - i);
    if (A[i] > i) {
      c++;
      big[A[i] - i]++;
    } else if (A[i] == i) {
      a++;
      big[0]++;
    } else {
      b++;
    }
  }
  int pos = n, id = 0;
  int ans_pos = 0;
  long long tmp = ans;
  while (--pos) {
    id++;
    tmp += b + a;
    tmp -= c;
    b = b + a;
    c = c - big[id];
    a = big[id];
    if (A[pos + 1] == n) {
      tmp += n - 2;
      int t = id + n - 1;
      if (t <= n) big[t]++;
      b--;
      c++;
    } else if (A[pos + 1] == 1) {
      b--;
      a++;
      tmp -= n;
    } else {
      b--;
      c++;
      int t = id + A[pos + 1] - 1;
      if (t <= n) big[t]++;
      tmp += -abs(A[pos + 1] - n - 1) + (A[pos + 1] - 1);
    }
    if (tmp <= ans) {
      ans = tmp;
      ans_pos = id;
    }
  }
  printf("%I64d %d\n", ans, ans_pos);
  return 0;
}

```