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;
vector<int> ev[1000099];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> p(n), pos(n);
  for (int(i) = (0); (i) < (int)(n); ++(i)) {
    cin >> p[i];
    p[i]--;
    pos[p[i]] = i;
    ev[n - i].push_back(p[i]);
    int k = (p[i] - i + n) % n;
    if (k != n - i) ev[k].push_back(p[i]);
  }
  long long sm = 0;
  int x = 0, y = 0;
  vector<int> state(n);
  for (int(i) = (0); (i) < (int)(n); ++(i)) {
    if (i < p[i])
      x++, state[p[i]] = -1;
    else
      y++, state[p[i]] = 1;
    sm += abs(i - p[i]);
  }
  long long mi = LLONG_MAX;
  int ans = -1;
  for (int(i) = (0); (i) < (int)(n); ++(i)) {
    if (i > 0) {
      for (auto &(num) : (ev[i])) {
        int nextPos = (pos[num] + i) % n;
        int prePos = (pos[num] + i - 1) % n;
        int nextState = nextPos < num ? -1 : 1;
        sm += abs(nextPos - num) - abs(prePos - num);
        if (state[num] == -1 && nextState == 1) {
          sm--;
          x--;
          y++;
          state[num] = 1;
        } else if (state[num] == 1 && nextState == -1) {
          sm++;
          x++;
          y--;
          state[num] = -1;
        } else if (state[num] == nextState) {
          if (state[num] == 1)
            sm--;
          else
            sm++;
        }
      }
      sm -= x;
      sm += y;
    }
    if (sm < mi) {
      ans = i;
      mi = sm;
    }
  }
  cout << mi << ' ' << ans << endl;
}
