#include <bits/stdc++.h>
using namespace std;
int n;
long long d[1000005], s[1000005];
void AddAP(long long a, long long x, long long i, long long j) {
  if (i == n) return;
  s[i] += a;
  s[j + 1] -= (a + x * (j - i));
  d[i + 1] += x;
  d[j + 1] -= x;
}
int main() {
  int min_i = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p < i) {
      AddAP(i - p, 1, 0, n - i);
      AddAP(p - 1, -1, n - i + 1, n - i + p);
      AddAP(0, 1, n - i + p, n - 1);
    } else {
      AddAP(p - i, -1, 0, p - i);
      AddAP(0, 1, p - i, n - i);
      AddAP(p - 1, -1, n - i + 1, n - 1);
    }
  }
  for (long long i = 1; i < n; i++) d[i] += d[i - 1];
  for (long long i = 1; i < n; i++) {
    s[i] += (s[i - 1] + d[i]);
    if (s[i] < s[min_i]) min_i = i;
  }
  printf("%lld %d\n", s[min_i], min_i);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 n;
long long d[1000005], s[1000005];
void AddAP(long long a, long long x, long long i, long long j) {
  if (i == n) return;
  s[i] += a;
  s[j + 1] -= (a + x * (j - i));
  d[i + 1] += x;
  d[j + 1] -= x;
}
int main() {
  int min_i = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p < i) {
      AddAP(i - p, 1, 0, n - i);
      AddAP(p - 1, -1, n - i + 1, n - i + p);
      AddAP(0, 1, n - i + p, n - 1);
    } else {
      AddAP(p - i, -1, 0, p - i);
      AddAP(0, 1, p - i, n - i);
      AddAP(p - 1, -1, n - i + 1, n - 1);
    }
  }
  for (long long i = 1; i < n; i++) d[i] += d[i - 1];
  for (long long i = 1; i < n; i++) {
    s[i] += (s[i - 1] + d[i]);
    if (s[i] < s[min_i]) min_i = i;
  }
  printf("%lld %d\n", s[min_i], min_i);
  return 0;
}

```