#include <bits/stdc++.h>
void solve() {
  int n;
  std::cin >> n;
  int64_t cursum = 0, pos = 0;
  std::vector<int> v(n + n);
  int nonneg = 0;
  int left = 0;
  std::multiset<int> neg;
  for (int i = 0; i < n; ++i) {
    std::cin >> v[i];
    v[i + n] = v[i];
    cursum += abs(v[i] - (i + 1));
    if (v[i] - (i + 1) >= 0) {
      ++nonneg;
    } else {
      neg.insert(-(v[i] - (i + 1)));
    }
  }
  int64_t smallest = cursum;
  for (int i = 1; i < n; ++i) {
    cursum -= v[i - 1] - 1;
    --nonneg;
    cursum += nonneg - int(neg.size());
    ++left;
    if (!neg.empty()) {
      if (*neg.begin() == left) {
        nonneg += neg.count(left);
        neg.erase(left);
      }
    }
    if (v[i + n - 1] - n < 0) {
      neg.insert(-(v[i + n - 1] - n) + left);
    } else {
      ++nonneg;
    }
    cursum += abs(v[i + n - 1] - n);
    if (cursum < smallest) {
      smallest = cursum;
      pos = i;
    }
  }
  std::cout << smallest << " " << (n - pos) % n;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  solve();
  return 0;
}

### 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>
void solve() {
  int n;
  std::cin >> n;
  int64_t cursum = 0, pos = 0;
  std::vector<int> v(n + n);
  int nonneg = 0;
  int left = 0;
  std::multiset<int> neg;
  for (int i = 0; i < n; ++i) {
    std::cin >> v[i];
    v[i + n] = v[i];
    cursum += abs(v[i] - (i + 1));
    if (v[i] - (i + 1) >= 0) {
      ++nonneg;
    } else {
      neg.insert(-(v[i] - (i + 1)));
    }
  }
  int64_t smallest = cursum;
  for (int i = 1; i < n; ++i) {
    cursum -= v[i - 1] - 1;
    --nonneg;
    cursum += nonneg - int(neg.size());
    ++left;
    if (!neg.empty()) {
      if (*neg.begin() == left) {
        nonneg += neg.count(left);
        neg.erase(left);
      }
    }
    if (v[i + n - 1] - n < 0) {
      neg.insert(-(v[i + n - 1] - n) + left);
    } else {
      ++nonneg;
    }
    cursum += abs(v[i + n - 1] - n);
    if (cursum < smallest) {
      smallest = cursum;
      pos = i;
    }
  }
  std::cout << smallest << " " << (n - pos) % n;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  solve();
  return 0;
}

```