#include <bits/stdc++.h>
using namespace std;
int n;
int a[1000001];
int pos[1000001];
long long last;
long long p[1000001];
long long z[1000001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    if (i > a[i]) {
      long long x = abs(a[i] - i);
      p[0] += x;
      ++z[0];
      int pos = n - i;
      --z[pos];
      ++pos;
      p[pos] -= x + pos - 1;
      if (a[i] > 1) {
        long long x = abs(1 - a[i]);
        --z[pos];
        p[pos] += x;
        pos += a[i] - 1;
        ++z[pos];
      }
      ++z[pos];
    } else {
      long long x = abs(a[i] - i);
      p[0] += x;
      --z[0];
      int ppos = a[i] - i;
      ++z[ppos];
      if (a[i] != n) {
        ++z[ppos];
        ppos = n - i;
        z[ppos]--;
        p[ppos + 1] -= n - a[i];
      }
      ++ppos;
      p[ppos] += abs(a[i] - 1);
      --z[ppos];
    }
  }
  long long ans = 1e18;
  long long k = 0;
  long long now = 0;
  for (int i = 0; i < n; ++i) {
    now += last + p[i];
    if (now < ans) {
      ans = now;
      k = i;
    }
    last += z[i];
  }
  cout << ans << " " << k;
}

### 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>
using namespace std;
int n;
int a[1000001];
int pos[1000001];
long long last;
long long p[1000001];
long long z[1000001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    if (i > a[i]) {
      long long x = abs(a[i] - i);
      p[0] += x;
      ++z[0];
      int pos = n - i;
      --z[pos];
      ++pos;
      p[pos] -= x + pos - 1;
      if (a[i] > 1) {
        long long x = abs(1 - a[i]);
        --z[pos];
        p[pos] += x;
        pos += a[i] - 1;
        ++z[pos];
      }
      ++z[pos];
    } else {
      long long x = abs(a[i] - i);
      p[0] += x;
      --z[0];
      int ppos = a[i] - i;
      ++z[ppos];
      if (a[i] != n) {
        ++z[ppos];
        ppos = n - i;
        z[ppos]--;
        p[ppos + 1] -= n - a[i];
      }
      ++ppos;
      p[ppos] += abs(a[i] - 1);
      --z[ppos];
    }
  }
  long long ans = 1e18;
  long long k = 0;
  long long now = 0;
  for (int i = 0; i < n; ++i) {
    now += last + p[i];
    if (now < ans) {
      ans = now;
      k = i;
    }
    last += z[i];
  }
  cout << ans << " " << k;
}

```