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 a[1000000 + 9];
long long d[1000000 + 9];
long long inst[1000000 + 9];
long long perm[1000000 + 9];
int n;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] >= i) {
      int x = a[i] - i;
      int l = 0, r = x;
      inst[l] += x + 1;
      perm[l] += -1;
      perm[r + 1] += 1;
      l = x + 1, r = n - i;
      perm[l] += 1;
      perm[r + 1] -= 1;
      inst[r + 1] -= (r - l + 1);
      x = a[i] - 1;
      l = n - i + 1, r = n - 1;
      inst[l] += x + 1;
      perm[l] += -1;
    } else {
      int x = i - a[i];
      int l = 0, r = n - i;
      inst[l] += x;
      perm[l + 1] += 1;
      perm[r + 1] -= 1;
      inst[r + 1] -= x + (r - l);
      x = a[i] - 1;
      l = n - i + 1, r = l + x;
      inst[l] += x + 1;
      perm[l] -= 1;
      perm[r + 1] += 1;
      l = r + 1, r = n - 1;
      perm[l] += 1;
    }
  }
  long long val = 0, inc = 0;
  for (int i = 0; i < n; i++) {
    val += inst[i];
    inc += perm[i];
    val += inc;
    d[i] = val;
  }
  long long minDev = d[0];
  int bestShift = 0;
  for (int i = 0; i < n; i++) {
    if (d[i] < minDev) {
      minDev = d[i];
      bestShift = i;
    }
  }
  cout << minDev << " " << bestShift << '\n';
  return 0;
}
