#include <bits/stdc++.h>
using namespace std;
struct event_t {
  int time;
  int change_incr_by;
  int change_sum_by;
  bool operator<(const event_t &other) const { return (time < other.time); }
};
const int max_N = 1e6 + 10;
int P[max_N];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", P + i);
  }
  vector<event_t> events;
  long long incr = 0, curr = 0;
  for (int i = 0; i < N; ++i) {
    int x = P[i] - 1;
    curr += abs(x - i);
    if (x <= i) {
      ++incr;
      events.push_back({N - i + x, 2, 0});
    } else {
      --incr;
      events.push_back({x - i, 2, 0});
    }
    events.push_back({N - 1 - i, -1, 0});
    events.push_back({N - i, -1, x - (N - 1 - x)});
  }
  sort(events.begin(), events.end());
  int i = 0;
  const long long inf = 1e17 + 42;
  long long min_dev = inf, point = -1;
  for (int k = 0; k < N; ++k) {
    while (i < events.size() && events[i].time == k) {
      incr += events[i].change_incr_by;
      curr += events[i].change_sum_by;
      ++i;
    }
    if (tie(min_dev, point) > tie(curr, k)) {
      tie(min_dev, point) = tie(curr, k);
    }
    curr += incr;
  }
  cout << min_dev << ' ' << point << endl;
}

### Prompt

Create a solution in Cpp for 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;
struct event_t {
  int time;
  int change_incr_by;
  int change_sum_by;
  bool operator<(const event_t &other) const { return (time < other.time); }
};
const int max_N = 1e6 + 10;
int P[max_N];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", P + i);
  }
  vector<event_t> events;
  long long incr = 0, curr = 0;
  for (int i = 0; i < N; ++i) {
    int x = P[i] - 1;
    curr += abs(x - i);
    if (x <= i) {
      ++incr;
      events.push_back({N - i + x, 2, 0});
    } else {
      --incr;
      events.push_back({x - i, 2, 0});
    }
    events.push_back({N - 1 - i, -1, 0});
    events.push_back({N - i, -1, x - (N - 1 - x)});
  }
  sort(events.begin(), events.end());
  int i = 0;
  const long long inf = 1e17 + 42;
  long long min_dev = inf, point = -1;
  for (int k = 0; k < N; ++k) {
    while (i < events.size() && events[i].time == k) {
      incr += events[i].change_incr_by;
      curr += events[i].change_sum_by;
      ++i;
    }
    if (tie(min_dev, point) > tie(curr, k)) {
      tie(min_dev, point) = tie(curr, k);
    }
    curr += incr;
  }
  cout << min_dev << ' ' << point << endl;
}

```