#include <bits/stdc++.h>
using namespace std;
long long cx[(int)1e6 + 1], c1[(int)1e6 + 1];
int main() {
  int n;
  cin >> n;
  vector<int> ps(n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &ps[i]);
    --ps[i];
  }
  auto add = [&](int b, int a, int fr, int to) {
    cx[fr] += a;
    c1[fr] += b;
    cx[to] -= a;
    c1[to] -= b;
  };
  for (int i = 0; i < (n); ++i) {
    int x = ps[i];
    int r1 = (x + n - i) % n;
    int r2 = n - i;
    if (i <= x) {
      add(x - i, -1, 0, r1);
      add(i - x, 1, r1, r2);
      add(r2 + x, -1, r2, n);
    } else if (i > x) {
      add(i - x, 1, 0, r2);
      add(r1, -1, r2, r1);
      add(-r1, 1, r1, n);
    }
  }
  long long a = 0, b = 0;
  vector<long long> cost(n);
  for (int i = 0; i < (n); ++i) {
    a += cx[i];
    b += c1[i];
    cost[i] = a * i + b;
  }
  pair<long long, long long> ans(1LL << 60, -1);
  for (int i = 0; i < (n); ++i) ans = min(ans, {cost[i], i});
  cout << ans.first << ' ' << ans.second << endl;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
long long cx[(int)1e6 + 1], c1[(int)1e6 + 1];
int main() {
  int n;
  cin >> n;
  vector<int> ps(n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &ps[i]);
    --ps[i];
  }
  auto add = [&](int b, int a, int fr, int to) {
    cx[fr] += a;
    c1[fr] += b;
    cx[to] -= a;
    c1[to] -= b;
  };
  for (int i = 0; i < (n); ++i) {
    int x = ps[i];
    int r1 = (x + n - i) % n;
    int r2 = n - i;
    if (i <= x) {
      add(x - i, -1, 0, r1);
      add(i - x, 1, r1, r2);
      add(r2 + x, -1, r2, n);
    } else if (i > x) {
      add(i - x, 1, 0, r2);
      add(r1, -1, r2, r1);
      add(-r1, 1, r1, n);
    }
  }
  long long a = 0, b = 0;
  vector<long long> cost(n);
  for (int i = 0; i < (n); ++i) {
    a += cx[i];
    b += c1[i];
    cost[i] = a * i + b;
  }
  pair<long long, long long> ans(1LL << 60, -1);
  for (int i = 0; i < (n); ++i) ans = min(ans, {cost[i], i});
  cout << ans.first << ' ' << ans.second << endl;
}

```