#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i]--;
  }
  multiset<int> leften;
  int rlen = 0;
  long long left_sum = 0;
  long long right_sum = 0;
  for (int i = 0; i < n; i++) {
    if (i < v[i]) {
      leften.insert(abs(v[i] - i));
      left_sum += abs(v[i] - i);
    } else {
      rlen += 1;
      right_sum += abs(v[i] - i);
    }
  }
  int best = 0;
  long long best_sum = left_sum + right_sum;
  reverse(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    if (best_sum > left_sum + right_sum) {
      best_sum = left_sum + right_sum;
      best = i;
    }
    left_sum -= leften.size();
    right_sum += rlen - 1;
    int ones = 0;
    while (!leften.empty() && (*leften.begin() - i) == 1) {
      leften.erase(leften.begin());
      rlen += 1;
      ones++;
    }
    int last = v[i];
    int last_prod = abs(last - (n - 1)) - i;
    rlen -= 1;
    if (last == 0) {
      rlen += 1;
    } else {
      leften.insert(abs(v[i]) + i + 1);
    }
    right_sum -= abs(last - (n - 1));
    left_sum += last;
  }
  cout << best_sum << " " << best;
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i]--;
  }
  multiset<int> leften;
  int rlen = 0;
  long long left_sum = 0;
  long long right_sum = 0;
  for (int i = 0; i < n; i++) {
    if (i < v[i]) {
      leften.insert(abs(v[i] - i));
      left_sum += abs(v[i] - i);
    } else {
      rlen += 1;
      right_sum += abs(v[i] - i);
    }
  }
  int best = 0;
  long long best_sum = left_sum + right_sum;
  reverse(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    if (best_sum > left_sum + right_sum) {
      best_sum = left_sum + right_sum;
      best = i;
    }
    left_sum -= leften.size();
    right_sum += rlen - 1;
    int ones = 0;
    while (!leften.empty() && (*leften.begin() - i) == 1) {
      leften.erase(leften.begin());
      rlen += 1;
      ones++;
    }
    int last = v[i];
    int last_prod = abs(last - (n - 1)) - i;
    rlen -= 1;
    if (last == 0) {
      rlen += 1;
    } else {
      leften.insert(abs(v[i]) + i + 1);
    }
    right_sum -= abs(last - (n - 1));
    left_sum += last;
  }
  cout << best_sum << " " << best;
}

```