#include <bits/stdc++.h>
using namespace std;
const int N = 1000007;
int p[N], d[N], len;
long long sum = 0, dsum;
struct Data {
  int id, val;
  int t;
  Data(int T = 0, int i = 0, int v = 0) : id(i), val(v), t(T) {}
} q[5 * N];
bool operator<(const Data& a, const Data& b) {
  if (a.t != b.t) return a.t < b.t;
  if (a.id != b.id) return a.id < b.id;
  return a.val < b.val;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  for (int i = 1; i <= n; i++) {
    if (p[i] <= i) {
      d[i] = 1;
      dsum++;
      sum += i - p[i];
      q[len++] = Data(n - i, i, -(n - p[i]) + p[i] - 1);
      q[len++] = Data(n - i + 1, i, -1);
      q[len++] = Data(n - i + p[i], i, 1);
    } else {
      d[i] = -1;
      dsum--;
      sum += p[i] - i;
      q[len++] = Data(p[i] - i, i, 1);
      q[len++] = Data(n - i, i, -(n - p[i]) + p[i] - 1);
      q[len++] = Data(n - i + 1, i, -1);
    }
  }
  sort(q, q + len);
  long long ans = sum;
  int j = 0, id = 0;
  for (int i = 0; i < n; i++) {
    if (sum < ans) {
      ans = sum;
      id = i;
    }
    while (q[j].t == i and j < len) {
      int val = q[j].val;
      int id = q[j].id;
      dsum -= d[id];
      dsum += val;
      d[id] = val;
      j++;
    }
    sum += dsum;
  }
  printf("%lld %d\n", ans, id);
  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>
using namespace std;
const int N = 1000007;
int p[N], d[N], len;
long long sum = 0, dsum;
struct Data {
  int id, val;
  int t;
  Data(int T = 0, int i = 0, int v = 0) : id(i), val(v), t(T) {}
} q[5 * N];
bool operator<(const Data& a, const Data& b) {
  if (a.t != b.t) return a.t < b.t;
  if (a.id != b.id) return a.id < b.id;
  return a.val < b.val;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  for (int i = 1; i <= n; i++) {
    if (p[i] <= i) {
      d[i] = 1;
      dsum++;
      sum += i - p[i];
      q[len++] = Data(n - i, i, -(n - p[i]) + p[i] - 1);
      q[len++] = Data(n - i + 1, i, -1);
      q[len++] = Data(n - i + p[i], i, 1);
    } else {
      d[i] = -1;
      dsum--;
      sum += p[i] - i;
      q[len++] = Data(p[i] - i, i, 1);
      q[len++] = Data(n - i, i, -(n - p[i]) + p[i] - 1);
      q[len++] = Data(n - i + 1, i, -1);
    }
  }
  sort(q, q + len);
  long long ans = sum;
  int j = 0, id = 0;
  for (int i = 0; i < n; i++) {
    if (sum < ans) {
      ans = sum;
      id = i;
    }
    while (q[j].t == i and j < len) {
      int val = q[j].val;
      int id = q[j].id;
      dsum -= d[id];
      dsum += val;
      d[id] = val;
      j++;
    }
    sum += dsum;
  }
  printf("%lld %d\n", ans, id);
  return 0;
}

```