#include <bits/stdc++.h>
const long long INF = 4e18L + 1;
const int IINF = 2e9 + 1;
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 1e6;
int n;
int perm[N + 1];
vector<int> events0[N + 1];
int events1[N + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long disc = 0;
  long long cnt_prev = 0, cnt_post = 0;
  for (long long i = (long long)1; i < (long long)n + 1; i++) {
    cin >> perm[i];
    disc += abs(i - perm[i]);
    events0[n + 1 - i].push_back(perm[i]);
    if (i < perm[i]) {
      events1[perm[i] - i]++;
      cnt_prev++;
    } else {
      events1[n - i + perm[i]]++;
      cnt_post++;
    }
  }
  long long res = disc, ans = 0;
  for (long long i = (long long)1; i < (long long)n; i++) {
    disc -= cnt_prev;
    disc += cnt_post;
    for (auto &p : events0[i]) {
      disc -= abs(p - (n + 1));
      disc += p - 1;
      cnt_post--;
      cnt_prev++;
    }
    cnt_prev -= events1[i];
    cnt_post += events1[i];
    if (disc < res) {
      res = disc;
      ans = i;
    }
  }
  cout << res << " " << ans << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
const long long INF = 4e18L + 1;
const int IINF = 2e9 + 1;
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 1e6;
int n;
int perm[N + 1];
vector<int> events0[N + 1];
int events1[N + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long disc = 0;
  long long cnt_prev = 0, cnt_post = 0;
  for (long long i = (long long)1; i < (long long)n + 1; i++) {
    cin >> perm[i];
    disc += abs(i - perm[i]);
    events0[n + 1 - i].push_back(perm[i]);
    if (i < perm[i]) {
      events1[perm[i] - i]++;
      cnt_prev++;
    } else {
      events1[n - i + perm[i]]++;
      cnt_post++;
    }
  }
  long long res = disc, ans = 0;
  for (long long i = (long long)1; i < (long long)n; i++) {
    disc -= cnt_prev;
    disc += cnt_post;
    for (auto &p : events0[i]) {
      disc -= abs(p - (n + 1));
      disc += p - 1;
      cnt_post--;
      cnt_prev++;
    }
    cnt_prev -= events1[i];
    cnt_post += events1[i];
    if (disc < res) {
      res = disc;
      ans = i;
    }
  }
  cout << res << " " << ans << "\n";
  return 0;
}

```