#include <bits/stdc++.h>
using namespace std;
long long dat[1000005];
struct node {
  long long val;
  node(long long a) { val = a; }
  bool operator<(const node& a) const { return val > a.val; }
};
int main() {
  int n, a;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &dat[i]);
    }
    long long ansSum = 99999999999999ll;
    long long tempSum = 0;
    int cnt = 0;
    priority_queue<node> q;
    for (int i = 1; i <= n; i++) {
      if (dat[i] > i) {
        q.push(dat[i] - i);
      } else
        cnt++;
      tempSum += abs(dat[i] - i);
    }
    int mov = 0;
    int ansmov = 0;
    for (int i = n; i > 0; i--) {
      if (tempSum < ansSum) {
        ansSum = tempSum;
        ansmov = mov;
      }
      tempSum += cnt - 1;
      tempSum -= q.size();
      tempSum += -(n - dat[i]) + (dat[i] - 1);
      mov++;
      if (dat[i] > 1) {
        q.push(dat[i] - 1 + mov);
        cnt--;
      }
      while (!q.empty() && q.top().val - mov <= 0) {
        q.pop();
        cnt++;
      }
    }
    printf("%lld %d\n", ansSum, ansmov);
  }
  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;
long long dat[1000005];
struct node {
  long long val;
  node(long long a) { val = a; }
  bool operator<(const node& a) const { return val > a.val; }
};
int main() {
  int n, a;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &dat[i]);
    }
    long long ansSum = 99999999999999ll;
    long long tempSum = 0;
    int cnt = 0;
    priority_queue<node> q;
    for (int i = 1; i <= n; i++) {
      if (dat[i] > i) {
        q.push(dat[i] - i);
      } else
        cnt++;
      tempSum += abs(dat[i] - i);
    }
    int mov = 0;
    int ansmov = 0;
    for (int i = n; i > 0; i--) {
      if (tempSum < ansSum) {
        ansSum = tempSum;
        ansmov = mov;
      }
      tempSum += cnt - 1;
      tempSum -= q.size();
      tempSum += -(n - dat[i]) + (dat[i] - 1);
      mov++;
      if (dat[i] > 1) {
        q.push(dat[i] - 1 + mov);
        cnt--;
      }
      while (!q.empty() && q.top().val - mov <= 0) {
        q.pop();
        cnt++;
      }
    }
    printf("%lld %d\n", ansSum, ansmov);
  }
  return 0;
}

```