#include <bits/stdc++.h>
using namespace std;
long long int n;
long long int nbMod[1000010];
vector<long long int> tab;
signed main() {
  cin >> n;
  long long int score = 0;
  long long int up = 0;
  for (long long int i = 1; i <= n; i++) {
    long long int a;
    cin >> a;
    tab.push_back(a);
    nbMod[(n + i - a) % n]++;
    if (a > i) up++;
    score += abs(i - a);
  }
  long long int bestScore = score;
  long long int iBest = -1;
  for (long long int i = 0; i < n - 1; i++) {
    long long int newScore = score;
    newScore -= abs(tab[i] - 1);
    newScore += abs(tab[i] - n);
    up += nbMod[i] - 1;
    newScore += up;
    newScore -= (n - 1 - up);
    if (newScore < bestScore) {
      bestScore = newScore;
      iBest = i;
    }
    score = newScore;
  }
  cout << bestScore << " " << (n - iBest - 1) % n << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
long long int n;
long long int nbMod[1000010];
vector<long long int> tab;
signed main() {
  cin >> n;
  long long int score = 0;
  long long int up = 0;
  for (long long int i = 1; i <= n; i++) {
    long long int a;
    cin >> a;
    tab.push_back(a);
    nbMod[(n + i - a) % n]++;
    if (a > i) up++;
    score += abs(i - a);
  }
  long long int bestScore = score;
  long long int iBest = -1;
  for (long long int i = 0; i < n - 1; i++) {
    long long int newScore = score;
    newScore -= abs(tab[i] - 1);
    newScore += abs(tab[i] - n);
    up += nbMod[i] - 1;
    newScore += up;
    newScore -= (n - 1 - up);
    if (newScore < bestScore) {
      bestScore = newScore;
      iBest = i;
    }
    score = newScore;
  }
  cout << bestScore << " " << (n - iBest - 1) % n << endl;
  return 0;
}

```