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 N = 1e6;
int n;
int p[N + 1];
int bit[N + 2];
inline int query(int idx) {
  int ret = 0;
  while (idx > 0) {
    ret += bit[idx];
    idx -= (idx & -idx);
  }
  return ret;
}
inline void bit_update(int idx, int val) {
  while (idx < N + 2) {
    bit[idx] += val;
    idx += (idx & -idx);
  }
}
void update(int l, int r, int val) {
  if (l > r) {
    return;
  };
  bit_update(l, val);
  bit_update(r + 1, -val);
}
void update2(int i) {
  ;
  update(1, p[i] - i, -1);
  update(p[i] - i + 1, p[i] - i + 1 + n - 1 - p[i] - 1, 1);
  update(p[i] - i + 1 + n - 1 - p[i] - 1 + 2, n - 1, -1);
}
void update1(int i) {
  ;
  update(1, n - 1 - i, 1);
  update(n - 1 - i + 2, n - 1 - i + 2 + p[i] - 1, -1);
  update(n - 1 - i + 2 + p[i] - 1 + 1, n - 1, 1);
}
void solve() {
  long long int total = 0;
  for (int i = 0, _stop = n - 1; i <= _stop; i++) {
    total += abs(i - p[i]);
    if (p[i] <= i) {
      update1(i);
    } else {
      update2(i);
    }
  }
  long long int ans = total;
  ;
  int shift = 0;
  int idx = n - 1;
  for (int i = 1, _stop = n - 1; i <= _stop; i++) {
    int q = query(i);
    int diff = p[idx] - abs(n - 1 - p[idx]);
    total += q + diff;
    ;
    if (ans > total) {
      ans = total;
      shift = i;
    }
    idx--;
  }
  cout << ans << " " << shift << '\n';
}
int main() {
  ;
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  {
    fill_n(bit, N + 2, 0);
    cin >> n;
    for (int i = 0, _stop = n - 1; i <= _stop; i++) {
      cin >> p[i];
      p[i]--;
    }
    solve();
  };
  return 0;
}
