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.
#include <bits/stdc++.h>
using namespace std;
int n, a[1100000], pos;
int zh, fu;
vector<int> f[1100000];
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > i)
      f[a[i] - i].push_back(i);
    else if (a[i] < i)
      f[n - i + a[i]].push_back(i);
    else
      f[0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    ans += (long long)abs(a[i] - i);
    if (a[i] - i < 0)
      fu++;
    else if (a[i] - i > 0)
      zh++;
  }
  long long now = ans;
  for (int i = 1; i < n; i++) {
    int last = n - i + 1;
    if (a[last] < n) fu--;
    now -= (long long)zh;
    now += (long long)fu;
    now += (long long)f[i - 1].size();
    if (a[n - i + 1] == n) now--;
    now -= (long long)abs(a[last] - n);
    now += (long long)(a[last] - 1);
    if (a[last] > 1) zh++;
    for (int j = 0; j < f[i - 1].size(); j++) {
      int x = a[f[i - 1][j]];
      int y = f[i - 1][j] + i;
      if (y == n + 1) continue;
      if (y > n) y -= n;
      if (x > y)
        zh++;
      else
        fu++;
    }
    for (int j = 0; j < f[i].size(); j++) {
      int x = a[f[i][j]];
      int y = f[i][j] + i - 1;
      if (y == n) continue;
      if (y > n) y -= n;
      if (x > y)
        zh--;
      else
        fu--;
    }
    if (now < ans) ans = now, pos = i;
  }
  cout << ans << " " << pos << endl;
  return 0;
}
