#include <bits/stdc++.h>
using namespace std;
int p[1000056], cnt[6000056];
int main() {
  int n;
  cin >> n;
  long long sum = 0, mn, po = 0, ne = 0, ans;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    sum += abs(p[i] - i);
    cnt[2000056 + p[i] - i]++;
    if (p[i] <= i)
      ne++;
    else
      po++;
  }
  mn = sum, ans = 0;
  for (int i = 1; i < n; i++) {
    sum += ne - po;
    sum += (p[n + 1 - i] - 1);
    sum -= (n + 1 - p[n + 1 - i]);
    if (sum < mn) {
      mn = sum, ans = i;
    }
    po -= cnt[2000056 + i];
    ne += cnt[2000056 + i];
    cnt[2000056 + p[n + 1 - i] + i - n - 1]--;
    cnt[2000056 + p[n + 1 - i] + i - 1]++;
    if (p[n + 1 - i] != 1) {
      po++;
      ne--;
    }
  }
  cout << mn << " " << ans;
}

### Prompt

Your challenge is to write 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;
int p[1000056], cnt[6000056];
int main() {
  int n;
  cin >> n;
  long long sum = 0, mn, po = 0, ne = 0, ans;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    sum += abs(p[i] - i);
    cnt[2000056 + p[i] - i]++;
    if (p[i] <= i)
      ne++;
    else
      po++;
  }
  mn = sum, ans = 0;
  for (int i = 1; i < n; i++) {
    sum += ne - po;
    sum += (p[n + 1 - i] - 1);
    sum -= (n + 1 - p[n + 1 - i]);
    if (sum < mn) {
      mn = sum, ans = i;
    }
    po -= cnt[2000056 + i];
    ne += cnt[2000056 + i];
    cnt[2000056 + p[n + 1 - i] + i - n - 1]--;
    cnt[2000056 + p[n + 1 - i] + i - 1]++;
    if (p[n + 1 - i] != 1) {
      po++;
      ne--;
    }
  }
  cout << mn << " " << ans;
}

```