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;
mt19937 rnd(rand());
long long inf = 1e9;
long long mod = 1e9 + 7;
long long big_inf = 1e18;
signed main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<long long> a(n);
  set<pair<long long, long long> > s1;
  long long s2_size = 0;
  vector<long long> set_val(n);
  long long my_sum = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    long long r = i + 1 - a[i];
    set_val[i] = r;
    if (r < 0)
      s1.insert({r, i});
    else
      s2_size++;
    my_sum += abs(i + 1 - a[i]);
  }
  long long j = n - 1;
  long long ans = my_sum;
  long long ans_num = 0;
  for (long long i = 0; i < n - 1; ++i) {
    my_sum -= abs(n - a[j]);
    my_sum += abs(1 - a[j]);
    my_sum -= s1.size();
    my_sum += s2_size - 1;
    if (my_sum < ans) {
      ans = my_sum;
      ans_num = i + 1;
    }
    while (s1.size() && s1.rbegin()->first + i + 1 >= 0) {
      s2_size++;
      s1.erase(*s1.rbegin());
    }
    if (a[j] > 1) {
      s2_size--;
      s1.insert({1 - a[j] - i - 1, j});
    }
    --j;
  }
  cout << ans << " " << ans_num;
  return 0;
}
