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;
const int MXN = 3e6 + 1;
long long N, A[MXN];
vector<int> chgs[MXN] = {};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  long long res = -(N - 1LL) * (N) / 2LL, val = 0;
  for (int x = 0; x < N; x++) cin >> A[x], A[x]--;
  long long ans = 0, idx = 0;
  long long add = 0, add_cnt = 0;
  for (int x = 0; x < N; x++)
    if (A[x] == N - 1) {
      val += A[x];
    } else if (A[x] >= x) {
      chgs[A[x] - x + 1].push_back(x);
      chgs[N - x].push_back(x);
      val += A[x];
    } else {
      chgs[N - x].push_back(x);
      chgs[N - x + A[x] + 1].push_back(x);
      val -= A[x];
      add += 2LL * x;
      add_cnt++;
    }
  ans = res + val + add;
  add += 2LL * add_cnt;
  for (int x = 1; x < N; x++) {
    for (int v : chgs[x])
      if (A[v] >= (v + x) % N)
        add -= 2LL * (v + x), add_cnt--, val += 2LL * A[v];
      else
        add += 2LL * (A[v] + 1), add_cnt++, val -= 2LL * A[v];
    if (res + val + add < ans) ans = res + val + add, idx = x;
    add += add_cnt * 2LL;
  }
  cout << ans << " " << idx << "\n";
}
