#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-6;
const int MX = 3333333;
const int P = 998244353;
int n;
int w[MX], num[MX];
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
    for (int i = 0; i <= 3 * n; i++) num[i] = 0;
    long long a = 0, b = 0, pre = 0;
    for (int i = 1; i <= n; i++) {
      num[2 * n + i - w[i]]++;
      pre += abs(i - w[i]);
      if (i - w[i] < 0) a++;
      if (i - w[i] >= 0) b++;
    }
    int t = n, id = n;
    long long ans = pre;
    for (int i = 2 * n; i > n + 1; i--, t--) {
      pre = pre - a + b;
      pre -= abs(n + 1 - w[t]);
      pre += abs(w[t] - 1);
      if (ans > pre) {
        ans = pre, id = t - 1;
      }
      b--;
      a++;
      num[i + n - w[t]]--;
      num[i - w[t]]++;
      a -= num[i - 1];
      b += num[i - 1];
    }
    cout << ans << " " << n - id << endl;
  }
  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;
const double eps = 1e-6;
const int MX = 3333333;
const int P = 998244353;
int n;
int w[MX], num[MX];
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
    for (int i = 0; i <= 3 * n; i++) num[i] = 0;
    long long a = 0, b = 0, pre = 0;
    for (int i = 1; i <= n; i++) {
      num[2 * n + i - w[i]]++;
      pre += abs(i - w[i]);
      if (i - w[i] < 0) a++;
      if (i - w[i] >= 0) b++;
    }
    int t = n, id = n;
    long long ans = pre;
    for (int i = 2 * n; i > n + 1; i--, t--) {
      pre = pre - a + b;
      pre -= abs(n + 1 - w[t]);
      pre += abs(w[t] - 1);
      if (ans > pre) {
        ans = pre, id = t - 1;
      }
      b--;
      a++;
      num[i + n - w[t]]--;
      num[i - w[t]]++;
      a -= num[i - 1];
      b += num[i - 1];
    }
    cout << ans << " " << n - id << endl;
  }
  return 0;
}

```