#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 5;
int arr[MAX];
long long sum3[MAX];
long long sum2[MAX];
long long sum1[MAX];
long long res[MAX];
void add(int l, int r, int st, int dif) {
  if (l > r) {
    return;
  } else {
    sum1[l] -= 1ll * l * 1ll * dif;
    sum1[r + 1] += 1ll * l * 1ll * dif;
    sum2[l] += st;
    sum2[r + 1] -= st;
    sum3[l] += dif;
    sum3[r + 1] -= dif;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (i >= arr[i]) {
      add(0, n - i, i - arr[i], 1);
      add(n - i + 1, n - i + arr[i], arr[i] - 1, -1);
      add(n - i + arr[i] + 1, n - 1, 1, 1);
    } else {
      add(0, arr[i] - i, arr[i] - i, -1);
      add(arr[i] - i + 1, n - i, 1, 1);
      add(n - i + 1, n - 1, arr[i] - 1, -1);
    }
  }
  long long sumc1 = 0, sumc2 = 0, sumc3 = 0;
  for (int i = 0; i < n; i++) {
    sumc1 += sum1[i];
    sumc2 += sum2[i];
    sumc3 += sum3[i];
    res[i] += sumc1 + sumc2 + (i * 1ll * sumc3);
  }
  long long bst1 = 0, bst2 = res[0];
  for (int i = 1; i < n; i++) {
    if (res[i] < bst2) {
      bst2 = res[i];
      bst1 = i;
    }
  }
  printf("%lld %lld", bst2, bst1);
}

### Prompt

Create a solution in cpp for 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;
const int MAX = 1e6 + 5;
int arr[MAX];
long long sum3[MAX];
long long sum2[MAX];
long long sum1[MAX];
long long res[MAX];
void add(int l, int r, int st, int dif) {
  if (l > r) {
    return;
  } else {
    sum1[l] -= 1ll * l * 1ll * dif;
    sum1[r + 1] += 1ll * l * 1ll * dif;
    sum2[l] += st;
    sum2[r + 1] -= st;
    sum3[l] += dif;
    sum3[r + 1] -= dif;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (i >= arr[i]) {
      add(0, n - i, i - arr[i], 1);
      add(n - i + 1, n - i + arr[i], arr[i] - 1, -1);
      add(n - i + arr[i] + 1, n - 1, 1, 1);
    } else {
      add(0, arr[i] - i, arr[i] - i, -1);
      add(arr[i] - i + 1, n - i, 1, 1);
      add(n - i + 1, n - 1, arr[i] - 1, -1);
    }
  }
  long long sumc1 = 0, sumc2 = 0, sumc3 = 0;
  for (int i = 0; i < n; i++) {
    sumc1 += sum1[i];
    sumc2 += sum2[i];
    sumc3 += sum3[i];
    res[i] += sumc1 + sumc2 + (i * 1ll * sumc3);
  }
  long long bst1 = 0, bst2 = res[0];
  for (int i = 1; i < n; i++) {
    if (res[i] < bst2) {
      bst2 = res[i];
      bst1 = i;
    }
  }
  printf("%lld %lld", bst2, bst1);
}

```