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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<long long> turns_positive_on(n);
  for (long long i = 0; i < n; i++) {
    turns_positive_on[a[i] - 1] = (a[i] - i - 1 + n) % n;
  }
  vector<vector<long long>> transition_map(n, vector<long long>({}));
  for (long long i = 0; i < n; i++) {
    transition_map[turns_positive_on[i]].push_back((i + 1));
  }
  long long curr = 0;
  for (long long i = 0; i < n; i++) curr += abs(a[i] - (i + 1));
  long long ans = curr;
  long long shiftID = 0;
  long long incr = 0;
  long long decr = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] <= (i + 1))
      incr++;
    else
      decr++;
  }
  long long lastPtr = (n - 1);
  for (long long shift = 1; shift < n; shift++) {
    long long last = a[lastPtr];
    curr += incr;
    curr -= decr;
    assert(incr);
    curr--;
    curr -= (abs(n - last));
    curr += (abs(1 - last));
    lastPtr--;
    incr--;
    decr++;
    decr -= (long long)(transition_map[shift].size());
    incr += (long long)(transition_map[shift].size());
    if (curr < ans) {
      ans = curr;
      shiftID = shift;
    }
  }
  cout << (ans) << ' ' << (shiftID) << ' ' << "\n";
  return 0;
}
