#include <bits/stdc++.h>
using namespace std;
int num1[1000001], num2[1000001];
int main() {
  int n;
  int add, sub, index;
  long long sum, re;
  while (scanf("%d", &n) != EOF) {
    sum = 0;
    add = 0;
    sub = 0;
    memset(num2, 0, sizeof(num2));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &num1[i]);
      sum += fabs(num1[i] - i);
      num2[(num1[i] - i + n) % n]++;
      if (num1[i] > i)
        sub++;
      else
        add++;
    }
    re = sum;
    index = 0;
    for (int i = 1; i < n; i++) {
      sum += add - sub - 1;
      sum += (num1[n - i + 1] - 1) - (n - num1[n - i + 1]);
      add = add + num2[i] - 1;
      sub = sub - num2[i] + 1;
      if (sum < re) {
        re = sum;
        index = i;
      }
    }
    printf("%lld %d\n", re, index);
  }
  return 0;
}

### Prompt

Your task is to create 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 num1[1000001], num2[1000001];
int main() {
  int n;
  int add, sub, index;
  long long sum, re;
  while (scanf("%d", &n) != EOF) {
    sum = 0;
    add = 0;
    sub = 0;
    memset(num2, 0, sizeof(num2));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &num1[i]);
      sum += fabs(num1[i] - i);
      num2[(num1[i] - i + n) % n]++;
      if (num1[i] > i)
        sub++;
      else
        add++;
    }
    re = sum;
    index = 0;
    for (int i = 1; i < n; i++) {
      sum += add - sub - 1;
      sum += (num1[n - i + 1] - 1) - (n - num1[n - i + 1]);
      add = add + num2[i] - 1;
      sub = sub - num2[i] + 1;
      if (sum < re) {
        re = sum;
        index = i;
      }
    }
    printf("%lld %d\n", re, index);
  }
  return 0;
}

```