#include <bits/stdc++.h>
using namespace std;
const long long int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
long long int p[MAXN];
long long int A[MAXN], B[MAXN];
void add(long long int l, long long int r, long long int diff,
         long long int first_term) {
  A[l] += first_term;
  B[l] += diff;
  A[r + 1] -= first_term + diff * (r - l);
  B[r] -= diff;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; ++i) cin >> p[i];
  for (int i = 1; i < n + 1; ++i) {
    if (p[i] >= i) {
      add(1, p[i] - i + 1, -1, p[i] - i);
      add(p[i] - i + 1, n - i + 1, 1, 0);
      add(n - i + 2, n, -1, p[i] - 1);
    } else {
      add(1, n - i + 1, 1, i - p[i]);
      add(n - i + 2, n - i + p[i] + 1, -1, p[i] - 1);
      add(n - i + p[i] + 1, n, 1, 0);
    }
  }
  for (int i = 1; i < n + 1; ++i)
    A[i] += B[i - 1], B[i] += B[i - 1], A[i] += A[i - 1];
  long long int res = LONG_LONG_MAX, pos = 0;
  for (int i = 1; i < n + 1; ++i) {
    if (A[i] < res) {
      res = A[i];
      pos = i;
    }
  }
  cout << A[pos] << " " << pos - 1 << endl;
}

### 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;
const long long int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
long long int p[MAXN];
long long int A[MAXN], B[MAXN];
void add(long long int l, long long int r, long long int diff,
         long long int first_term) {
  A[l] += first_term;
  B[l] += diff;
  A[r + 1] -= first_term + diff * (r - l);
  B[r] -= diff;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; ++i) cin >> p[i];
  for (int i = 1; i < n + 1; ++i) {
    if (p[i] >= i) {
      add(1, p[i] - i + 1, -1, p[i] - i);
      add(p[i] - i + 1, n - i + 1, 1, 0);
      add(n - i + 2, n, -1, p[i] - 1);
    } else {
      add(1, n - i + 1, 1, i - p[i]);
      add(n - i + 2, n - i + p[i] + 1, -1, p[i] - 1);
      add(n - i + p[i] + 1, n, 1, 0);
    }
  }
  for (int i = 1; i < n + 1; ++i)
    A[i] += B[i - 1], B[i] += B[i - 1], A[i] += A[i - 1];
  long long int res = LONG_LONG_MAX, pos = 0;
  for (int i = 1; i < n + 1; ++i) {
    if (A[i] < res) {
      res = A[i];
      pos = i;
    }
  }
  cout << A[pos] << " " << pos - 1 << endl;
}

```