#include <bits/stdc++.h>
using namespace std;
multiset<int> b;
multiset<int>::iterator it;
int sia = 0;
long long ansa = 0, ansb = 0;
stack<int> sta;
int k[1000005];
int main() {
  int n, ansk = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", k + i);
    if (k[i] > i) {
      b.insert(k[i] - i);
      ansb += k[i] - i;
    } else {
      ++sia;
      ansa += i - k[i];
    }
  }
  long long mi = ansa + ansb;
  int co = 1;
  for (int i = n; i > 1; --i) {
    ansb -= b.size();
    ansa += sia;
    int t = b.size();
    it = b.upper_bound(co);
    b.erase(b.begin(), it);
    t -= b.size();
    sia += t;
    if (k[i] > n + 1) {
      b.erase(b.find(k[i] - n - 1 + co));
      ansb -= k[i] - n + 1;
    } else {
      --sia;
      ansa -= n + 1 - k[i];
    }
    if (k[i] > 1) {
      b.insert(k[i] - 1 + co);
      ansb += k[i] - 1;
    } else {
      ++sia;
      ansa += 1 - k[i];
    }
    ++co;
    if (mi > ansa + ansb) {
      mi = ansa + ansb;
      ansk = co - 1;
    }
  }
  cout << mi << ' ' << ansk << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
multiset<int> b;
multiset<int>::iterator it;
int sia = 0;
long long ansa = 0, ansb = 0;
stack<int> sta;
int k[1000005];
int main() {
  int n, ansk = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", k + i);
    if (k[i] > i) {
      b.insert(k[i] - i);
      ansb += k[i] - i;
    } else {
      ++sia;
      ansa += i - k[i];
    }
  }
  long long mi = ansa + ansb;
  int co = 1;
  for (int i = n; i > 1; --i) {
    ansb -= b.size();
    ansa += sia;
    int t = b.size();
    it = b.upper_bound(co);
    b.erase(b.begin(), it);
    t -= b.size();
    sia += t;
    if (k[i] > n + 1) {
      b.erase(b.find(k[i] - n - 1 + co));
      ansb -= k[i] - n + 1;
    } else {
      --sia;
      ansa -= n + 1 - k[i];
    }
    if (k[i] > 1) {
      b.insert(k[i] - 1 + co);
      ansb += k[i] - 1;
    } else {
      ++sia;
      ansa += 1 - k[i];
    }
    ++co;
    if (mi > ansa + ansb) {
      mi = ansa + ansb;
      ansk = co - 1;
    }
  }
  cout << mi << ' ' << ansk << endl;
  return 0;
}

```