#include <bits/stdc++.h>
const int N = 2000005;
const int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
int n, p[N];
long long sk[N], sb[N];
using namespace std;
void add(int l, int r, int k, int y) {
  int b = y - k * l;
  sk[l] += k, sb[l] += b, sk[r + 1] -= k, sb[r + 1] -= b;
}
int main() {
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) scanf("%d", &p[i]), p[i]--;
  for (i = 0; i < n; i++)
    if (p[i] >= i) {
      int mid = p[i] - i;
      add(0, mid, -1, mid);
      add(mid + 1, n - 1 - i, 1, 1);
      add(n - i, n - 1, -1, p[i]);
    } else {
      add(0, n - 1 - i, 1, i - p[i]);
      add(n - i, n - i + p[i], -1, p[i]);
      add(n - i + p[i] + 1, n - 1, 1, 1);
    }
  long long ans = 1ll << 60;
  int pos = 0;
  for (i = 0; i < n; i++) {
    sk[i] += sk[i - 1], sb[i] += sb[i - 1];
    long long val = sk[i] * i + sb[i];
    if (val < ans) ans = val, pos = i;
  }
  cout << ans << " " << pos;
  return 0;
}

### 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>
const int N = 2000005;
const int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
int n, p[N];
long long sk[N], sb[N];
using namespace std;
void add(int l, int r, int k, int y) {
  int b = y - k * l;
  sk[l] += k, sb[l] += b, sk[r + 1] -= k, sb[r + 1] -= b;
}
int main() {
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) scanf("%d", &p[i]), p[i]--;
  for (i = 0; i < n; i++)
    if (p[i] >= i) {
      int mid = p[i] - i;
      add(0, mid, -1, mid);
      add(mid + 1, n - 1 - i, 1, 1);
      add(n - i, n - 1, -1, p[i]);
    } else {
      add(0, n - 1 - i, 1, i - p[i]);
      add(n - i, n - i + p[i], -1, p[i]);
      add(n - i + p[i] + 1, n - 1, 1, 1);
    }
  long long ans = 1ll << 60;
  int pos = 0;
  for (i = 0; i < n; i++) {
    sk[i] += sk[i - 1], sb[i] += sb[i - 1];
    long long val = sk[i] * i + sb[i];
    if (val < ans) ans = val, pos = i;
  }
  cout << ans << " " << pos;
  return 0;
}

```