#include <bits/stdc++.h>
using namespace std;
const int MAX = 1000000;
int p[MAX + 1];
struct event {
  long long da, db, i;
  int tme;
  bool operator<(const event& e) const { return tme < e.tme; }
};
int main() {
  int n;
  scanf("%d", &n);
  long long a = 0, b = 0;
  vector<event> tme;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i] >= i) {
      b += p[i] - i;
      a--;
      tme.push_back({2, 2 * (i - p[i]), i, p[i] - i});
    } else if (p[i] < i) {
      b += i - p[i];
      a++;
      tme.push_back({2, -2 * (p[i] + (n - i)), i, p[i] - i + n});
    }
    tme.push_back({-2, -i + (n - i) + 2 * p[i], i, n - i});
  }
  sort(tme.begin(), tme.end());
  int ptr = 0;
  long long best_deviation = 0x3f3f3f3f3f3f3f3fLL;
  int best_offset = -1;
  for (int i = 0; i < n; i++) {
    long long deviation = a * i + b;
    if (deviation < best_deviation) {
      best_offset = i;
      best_deviation = deviation;
    }
    while (ptr < tme.size() && tme[ptr].tme == i) {
      a += tme[ptr].da;
      b += tme[ptr].db;
      ptr++;
    }
  }
  cout << best_deviation << " " << best_offset << endl;
}

### Prompt

Generate 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;
const int MAX = 1000000;
int p[MAX + 1];
struct event {
  long long da, db, i;
  int tme;
  bool operator<(const event& e) const { return tme < e.tme; }
};
int main() {
  int n;
  scanf("%d", &n);
  long long a = 0, b = 0;
  vector<event> tme;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i] >= i) {
      b += p[i] - i;
      a--;
      tme.push_back({2, 2 * (i - p[i]), i, p[i] - i});
    } else if (p[i] < i) {
      b += i - p[i];
      a++;
      tme.push_back({2, -2 * (p[i] + (n - i)), i, p[i] - i + n});
    }
    tme.push_back({-2, -i + (n - i) + 2 * p[i], i, n - i});
  }
  sort(tme.begin(), tme.end());
  int ptr = 0;
  long long best_deviation = 0x3f3f3f3f3f3f3f3fLL;
  int best_offset = -1;
  for (int i = 0; i < n; i++) {
    long long deviation = a * i + b;
    if (deviation < best_deviation) {
      best_offset = i;
      best_deviation = deviation;
    }
    while (ptr < tme.size() && tme[ptr].tme == i) {
      a += tme[ptr].da;
      b += tme[ptr].db;
      ptr++;
    }
  }
  cout << best_deviation << " " << best_offset << endl;
}

```