#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1000006;
const long long Inf = 9000000000000000000ll;
int n;
int p[Maxn];
long long A[Maxn], B[Maxn];
void Insert(int l, int r, int val, int delt) {
  A[l] += val;
  B[l] += delt;
  A[r + 1] -= (val + delt * (r + 1 - l));
  B[r + 1] -= delt;
}
void smartInsert(int l, int r, int ind, int val, int delt) {
  if (ind <= l)
    Insert(l - ind, r - ind, val, delt);
  else if (ind > r)
    Insert(l - ind + n, r - ind + n, val, delt);
  else {
    Insert(l - ind + n, n - 1, val, delt);
    val += (ind - l) * delt;
    Insert(0, r - ind, val, delt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    smartInsert(p[i], n - 1, i, 0, 1);
    smartInsert(0, p[i], i, p[i], -1);
  }
  long long res = Inf;
  int wth;
  long long val = 0, delt = 0;
  for (int i = 0; i < n; i++) {
    val += A[i];
    if (val < res) {
      res = val;
      wth = i;
    }
    delt += B[i];
    val += delt;
  }
  printf("%I64d %d\n", res, wth);
  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;
const int Maxn = 1000006;
const long long Inf = 9000000000000000000ll;
int n;
int p[Maxn];
long long A[Maxn], B[Maxn];
void Insert(int l, int r, int val, int delt) {
  A[l] += val;
  B[l] += delt;
  A[r + 1] -= (val + delt * (r + 1 - l));
  B[r + 1] -= delt;
}
void smartInsert(int l, int r, int ind, int val, int delt) {
  if (ind <= l)
    Insert(l - ind, r - ind, val, delt);
  else if (ind > r)
    Insert(l - ind + n, r - ind + n, val, delt);
  else {
    Insert(l - ind + n, n - 1, val, delt);
    val += (ind - l) * delt;
    Insert(0, r - ind, val, delt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    smartInsert(p[i], n - 1, i, 0, 1);
    smartInsert(0, p[i], i, p[i], -1);
  }
  long long res = Inf;
  int wth;
  long long val = 0, delt = 0;
  for (int i = 0; i < n; i++) {
    val += A[i];
    if (val < res) {
      res = val;
      wth = i;
    }
    delt += B[i];
    val += delt;
  }
  printf("%I64d %d\n", res, wth);
  return 0;
}

```