#include <bits/stdc++.h>
using namespace std;
const double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int MR = 1e6 + 10;
int p[MR];
int main() {
  int n;
  scanf("%d", &n);
  pair<long long, int> add(0, 0), sub(0, 0);
  vector<int> aM(n, 0);
  int zero = 0;
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &p[i]);
    p[i]--;
    if (p[i] - i > 0) {
      add.first += p[i] - i;
      add.second++;
      aM[p[i] - i]++;
    }
    if (p[i] - i < 0) {
      sub.first += p[i] - i;
      sub.second++;
    }
    if (p[i] == i) zero++;
  }
  pair<long long, int> res = make_pair(add.first - sub.first, 0);
  int last = n - 1;
  int sh = 0;
  for (int i = (n)-1; i >= (1); --i) {
    if (p[i] - last > 0) {
      add.first -= p[i] - last;
      add.second--;
    }
    if (p[i] - i > 0) aM[p[i] - i]--;
    if (p[i] - last < 0) {
      sub.first -= p[i] - last;
      sub.second--;
    }
    if (p[i] == last) zero--;
    sh++;
    aM.push_back(0);
    add.first -= add.second;
    add.second -= aM[sh];
    sub.first -= sub.second;
    sub.first -= zero;
    sub.second += zero;
    zero = aM[sh];
    if (p[i] > 0) {
      add.first += p[i];
      add.second++;
      aM[sh + p[i]]++;
    } else
      zero++;
    res = min(res, make_pair(add.first - sub.first, n - i));
  }
  printf("%lld %d\n", res.first, res.second);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int MR = 1e6 + 10;
int p[MR];
int main() {
  int n;
  scanf("%d", &n);
  pair<long long, int> add(0, 0), sub(0, 0);
  vector<int> aM(n, 0);
  int zero = 0;
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &p[i]);
    p[i]--;
    if (p[i] - i > 0) {
      add.first += p[i] - i;
      add.second++;
      aM[p[i] - i]++;
    }
    if (p[i] - i < 0) {
      sub.first += p[i] - i;
      sub.second++;
    }
    if (p[i] == i) zero++;
  }
  pair<long long, int> res = make_pair(add.first - sub.first, 0);
  int last = n - 1;
  int sh = 0;
  for (int i = (n)-1; i >= (1); --i) {
    if (p[i] - last > 0) {
      add.first -= p[i] - last;
      add.second--;
    }
    if (p[i] - i > 0) aM[p[i] - i]--;
    if (p[i] - last < 0) {
      sub.first -= p[i] - last;
      sub.second--;
    }
    if (p[i] == last) zero--;
    sh++;
    aM.push_back(0);
    add.first -= add.second;
    add.second -= aM[sh];
    sub.first -= sub.second;
    sub.first -= zero;
    sub.second += zero;
    zero = aM[sh];
    if (p[i] > 0) {
      add.first += p[i];
      add.second++;
      aM[sh + p[i]]++;
    } else
      zero++;
    res = min(res, make_pair(add.first - sub.first, n - i));
  }
  printf("%lld %d\n", res.first, res.second);
  return 0;
}

```