#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int a[2 * N];
const long long INF = 1e15;
int n;
long long p[2 * N];
void add(int l, int r, long long val) {
  if (l > r) return;
  a[l] += val;
  a[r + 1] -= val;
}
void stress() {
  for (int i = n + 1; i <= n + n; i++) p[i] = p[i - n];
  long long ans = INF, anspos = -1;
  for (int i = 1; i <= n; i++) {
    long long int tans = 0;
    for (int j = 1; j <= n; j++) {
      tans += abs(p[i + j - 1] - j);
    }
    cout << tans << " ";
    if (tans < ans) {
      ans = tans;
      anspos = i;
    }
  }
  cout << endl;
  if (anspos == 1)
    anspos = 0;
  else
    anspos = (n - anspos + 1);
  cout << "stress " << ans << " " << anspos << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  memset(a, 0, sizeof(a));
  long long val = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i] >= i) {
      val += (p[i] - i);
      add(1, p[i] - i, -1);
      add(p[i] - i + 1, n - i, +1);
      add(n - i + 1, n - i + 1, 2 * p[i] - n - 1);
      add(n - i + 2, n - 1, -1);
    } else {
      val += (i - p[i]);
      add(1, n - i, +1);
      add(n - i + 1, n - i + 1, 2 * p[i] - 1 - n);
      add(n - i + 2, n - i + p[i], -1);
      add(n - i + p[i] + 1, n - 1, 1);
    }
  }
  long long ans = val, anspos = 0;
  for (int i = 1; i < n; i++) a[i] += a[i - 1];
  for (int i = 1; i < n; i++) {
    val = val + a[i];
    if (val < ans) {
      ans = val;
      anspos = i;
    }
  }
  cout << ans << " " << anspos << endl;
  return 0;
}

### 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 N = 1e6 + 5;
int a[2 * N];
const long long INF = 1e15;
int n;
long long p[2 * N];
void add(int l, int r, long long val) {
  if (l > r) return;
  a[l] += val;
  a[r + 1] -= val;
}
void stress() {
  for (int i = n + 1; i <= n + n; i++) p[i] = p[i - n];
  long long ans = INF, anspos = -1;
  for (int i = 1; i <= n; i++) {
    long long int tans = 0;
    for (int j = 1; j <= n; j++) {
      tans += abs(p[i + j - 1] - j);
    }
    cout << tans << " ";
    if (tans < ans) {
      ans = tans;
      anspos = i;
    }
  }
  cout << endl;
  if (anspos == 1)
    anspos = 0;
  else
    anspos = (n - anspos + 1);
  cout << "stress " << ans << " " << anspos << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  memset(a, 0, sizeof(a));
  long long val = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i] >= i) {
      val += (p[i] - i);
      add(1, p[i] - i, -1);
      add(p[i] - i + 1, n - i, +1);
      add(n - i + 1, n - i + 1, 2 * p[i] - n - 1);
      add(n - i + 2, n - 1, -1);
    } else {
      val += (i - p[i]);
      add(1, n - i, +1);
      add(n - i + 1, n - i + 1, 2 * p[i] - 1 - n);
      add(n - i + 2, n - i + p[i], -1);
      add(n - i + p[i] + 1, n - 1, 1);
    }
  }
  long long ans = val, anspos = 0;
  for (int i = 1; i < n; i++) a[i] += a[i - 1];
  for (int i = 1; i < n; i++) {
    val = val + a[i];
    if (val < ans) {
      ans = val;
      anspos = i;
    }
  }
  cout << ans << " " << anspos << endl;
  return 0;
}

```