#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
long long p[maxn], a[maxn], d[maxn];
void add(int b, int k, int l, int r) {
  if (l > r) return;
  a[l] += b;
  a[r + 1] -= b;
  d[l] += k;
  d[r] -= k;
  a[r + 1] -= 1LL * k * (r - l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i + 1];
  }
  memset(a, 0, sizeof a);
  memset(d, 0, sizeof d);
  for (int k = 1; k <= n; ++k) {
    long long x = p[k];
    if (x >= k) {
      add(x - k, -1, 0, x - k - 1);
      add(0, 1, x - k, n - k);
      add(x - 1, -1, n - k + 1, n - 1);
    } else {
      add(k - x, 1, 0, n - k);
      add(x - 1, -1, n - k + 1, x + n - k);
      add(1, 1, x + n - k + 1, n - 1);
    }
  }
  for (int i = 1; i < n; ++i) {
    d[i] += d[i - 1];
    a[i] += d[i - 1];
    a[i] += a[i - 1];
  }
  long long ans = 1e17, dev = -1;
  for (int i = 0; i < n; ++i) {
    if (a[i] < ans) {
      ans = a[i];
      dev = i;
    }
  }
  cout << ans << ' ' << dev << '\n';
}

### Prompt

Please formulate 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>
using namespace std;
const int maxn = 1e6 + 10;
long long p[maxn], a[maxn], d[maxn];
void add(int b, int k, int l, int r) {
  if (l > r) return;
  a[l] += b;
  a[r + 1] -= b;
  d[l] += k;
  d[r] -= k;
  a[r + 1] -= 1LL * k * (r - l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i + 1];
  }
  memset(a, 0, sizeof a);
  memset(d, 0, sizeof d);
  for (int k = 1; k <= n; ++k) {
    long long x = p[k];
    if (x >= k) {
      add(x - k, -1, 0, x - k - 1);
      add(0, 1, x - k, n - k);
      add(x - 1, -1, n - k + 1, n - 1);
    } else {
      add(k - x, 1, 0, n - k);
      add(x - 1, -1, n - k + 1, x + n - k);
      add(1, 1, x + n - k + 1, n - 1);
    }
  }
  for (int i = 1; i < n; ++i) {
    d[i] += d[i - 1];
    a[i] += d[i - 1];
    a[i] += a[i - 1];
  }
  long long ans = 1e17, dev = -1;
  for (int i = 0; i < n; ++i) {
    if (a[i] < ans) {
      ans = a[i];
      dev = i;
    }
  }
  cout << ans << ' ' << dev << '\n';
}

```