#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int number;
  std::cin >> number;
  std::vector<int64_t> diff2(number);
  auto const diffAdd = [&](int index, int64_t value) {
    if (index >= (int)diff2.size()) return;
    diff2[index] += value;
  };
  auto const add = [&](int left, int right, int64_t start, int64_t change) {
    assert(left < right);
    assert(right <= number);
    diffAdd(left, start);
    diffAdd(left + 1, -start);
    diffAdd(right, -start);
    diffAdd(right + 1, start);
    diffAdd(left + 1, change);
    diffAdd(right, -change);
    diffAdd(right, -(right - left - 1) * change);
    diffAdd(right + 1, (right - left - 1) * change);
  };
  auto const addAbs = [&](int left, int count, int start) {
    if (start < 0) {
      auto delta = std::min(count, -start);
      add(left, left + delta, -start, -1);
      start += delta;
      left += delta;
      count -= delta;
    }
    if (count == 0) return;
    assert(start >= 0);
    add(left, left + count, start, 1);
  };
  for (int i = 0; i < number; ++i) {
    int p;
    std::cin >> p;
    addAbs(0, number - i, i + 1 - p);
    if (i != 0) addAbs(number - i, i, 1 - p);
  }
  std::partial_sum(begin(diff2), end(diff2), diff2.begin());
  std::partial_sum(begin(diff2), end(diff2), diff2.begin());
  auto const iter = std::min_element(begin(diff2), end(diff2));
  std::cout << *iter << ' ' << iter - diff2.begin() << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int number;
  std::cin >> number;
  std::vector<int64_t> diff2(number);
  auto const diffAdd = [&](int index, int64_t value) {
    if (index >= (int)diff2.size()) return;
    diff2[index] += value;
  };
  auto const add = [&](int left, int right, int64_t start, int64_t change) {
    assert(left < right);
    assert(right <= number);
    diffAdd(left, start);
    diffAdd(left + 1, -start);
    diffAdd(right, -start);
    diffAdd(right + 1, start);
    diffAdd(left + 1, change);
    diffAdd(right, -change);
    diffAdd(right, -(right - left - 1) * change);
    diffAdd(right + 1, (right - left - 1) * change);
  };
  auto const addAbs = [&](int left, int count, int start) {
    if (start < 0) {
      auto delta = std::min(count, -start);
      add(left, left + delta, -start, -1);
      start += delta;
      left += delta;
      count -= delta;
    }
    if (count == 0) return;
    assert(start >= 0);
    add(left, left + count, start, 1);
  };
  for (int i = 0; i < number; ++i) {
    int p;
    std::cin >> p;
    addAbs(0, number - i, i + 1 - p);
    if (i != 0) addAbs(number - i, i, 1 - p);
  }
  std::partial_sum(begin(diff2), end(diff2), diff2.begin());
  std::partial_sum(begin(diff2), end(diff2), diff2.begin());
  auto const iter = std::min_element(begin(diff2), end(diff2));
  std::cout << *iter << ' ' << iter - diff2.begin() << '\n';
}

```