#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
const int MAX = 1e6 + 10;
int s[MAX];
int p[MAX];
vi flip[MAX];
ll best = 1e14;
int res, canti, n;
ll bigsum, bigsum2, smallsum, smallsum2;
int red(int x) {
  if (x > n) x -= n;
  return x;
}
ll shift(int i) {
  bigsum += canti;
  smallsum += n - canti;
  if (s[n - i + 1])
    bigsum -= n;
  else
    smallsum -= n;
  for (int x : flip[i]) {
    if (s[x]) {
      canti--;
      bigsum -= red(x + i);
      smallsum += red(x + i);
      bigsum2 -= p[x];
      smallsum2 += p[x];
    } else {
      canti++;
      bigsum += red(x + i);
      smallsum -= red(x + i);
      bigsum2 += p[x];
      smallsum2 -= p[x];
    }
    s[x] = (s[x] + 1) % 2;
  }
  return bigsum2 - bigsum + smallsum - smallsum2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] >= i) {
      s[i] = 1;
      canti++;
      bigsum += i;
      bigsum2 += p[i];
      flip[p[i] - i + 1].push_back(i);
      flip[n - i + 1].push_back(i);
    } else {
      smallsum += i;
      smallsum2 += p[i];
      flip[n - i + 1].push_back(i);
      if (i - 1 > p[i]) flip[n - i + p[i] + 1].push_back(i);
    }
  }
  best = bigsum2 - bigsum + smallsum - smallsum2;
  for (int i = 1; i < n; i++) {
    ll x = shift(i);
    if (x < best) {
      best = x;
      res = i;
    }
  }
  cout << best << " " << res << "\n";
}

### 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;
using ll = long long;
using vi = vector<int>;
const int MAX = 1e6 + 10;
int s[MAX];
int p[MAX];
vi flip[MAX];
ll best = 1e14;
int res, canti, n;
ll bigsum, bigsum2, smallsum, smallsum2;
int red(int x) {
  if (x > n) x -= n;
  return x;
}
ll shift(int i) {
  bigsum += canti;
  smallsum += n - canti;
  if (s[n - i + 1])
    bigsum -= n;
  else
    smallsum -= n;
  for (int x : flip[i]) {
    if (s[x]) {
      canti--;
      bigsum -= red(x + i);
      smallsum += red(x + i);
      bigsum2 -= p[x];
      smallsum2 += p[x];
    } else {
      canti++;
      bigsum += red(x + i);
      smallsum -= red(x + i);
      bigsum2 += p[x];
      smallsum2 -= p[x];
    }
    s[x] = (s[x] + 1) % 2;
  }
  return bigsum2 - bigsum + smallsum - smallsum2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] >= i) {
      s[i] = 1;
      canti++;
      bigsum += i;
      bigsum2 += p[i];
      flip[p[i] - i + 1].push_back(i);
      flip[n - i + 1].push_back(i);
    } else {
      smallsum += i;
      smallsum2 += p[i];
      flip[n - i + 1].push_back(i);
      if (i - 1 > p[i]) flip[n - i + p[i] + 1].push_back(i);
    }
  }
  best = bigsum2 - bigsum + smallsum - smallsum2;
  for (int i = 1; i < n; i++) {
    ll x = shift(i);
    if (x < best) {
      best = x;
      res = i;
    }
  }
  cout << best << " " << res << "\n";
}

```