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;
struct KK {
  vector<long long> v;
  KK(int sz) { v = vector<long long>(sz, 0); }
  void add(int l, int r, long long x) {
    v[l] += x;
    if (r < (int)v.size()) v[r] -= x;
  }
  vector<long long> getAns() {
    vector<long long> res = {v[0]};
    for (int i = 1; i < (int)v.size(); i++) {
      res.push_back(res.back() + v[i]);
    }
    return res;
  }
};
struct SEG {
  int l, r, val;
  SEG(int l = 0, int r = 0, int val = 0) : l(l), r(r), val(val) {}
};
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int &x : a) {
    cin >> x;
    x--;
  }
  long long ans = 0;
  KK kk(n);
  for (int i = 0; i < n; i++) {
    int val = a[i];
    int pos = i;
    ans += abs(i - a[i]);
    vector<SEG> segs = {SEG(0, val, -1), SEG(val, n - 1, 1),
                        SEG(n - 1, n, val + val + 1 - n)};
    int dx = -pos;
    for (const auto seg : segs) {
      int l = seg.l + dx;
      int r = seg.r + dx;
      if (l < 0) kk.add(n + l, n + min(0, r), seg.val);
      if (r > 0) kk.add(max(l, 0), min(r, n), seg.val);
    }
  }
  auto dx = kk.getAns();
  int pos = 0;
  long long cur = ans;
  for (int i = 0; i < n - 1; i++) {
    cur += dx[i];
    if (cur < ans) {
      ans = cur;
      pos = i + 1;
    }
  }
  cout << ans << " " << pos << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
}
