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;
int n;
int tab[1000001];
vector<int> V[3000001];
long long m = 0, p = 0, res = 0, tim = 0;
long long m_amount = 0, p_amount = 0;
bool ifmin[1000001];
void read_and_prepare() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> tab[i];
    if (tab[i] - i > 0) {
      p_amount++;
      p += tab[i] - i;
      V[tab[i] - i].push_back(i);
    } else {
      ifmin[i] = 1;
      m_amount++;
      m += i - tab[i];
    }
  }
  res = m + p;
}
void shifts() {
  for (int i = 1; i < n; i++) {
    int id = n - i + 1;
    m -= n - tab[id];
    p += tab[id] - 1;
    m_amount--;
    ifmin[id] = 0;
    p -= p_amount;
    m += m_amount;
    p_amount++;
    V[i + tab[id] - 1].push_back(id);
    for (unsigned int j = 0; j < V[i].size(); j++) {
      id = V[i][j];
      if (!ifmin[id]) {
        ifmin[id] = 1;
        p_amount--;
        m_amount++;
      }
    }
    if (m + p < res) {
      res = m + p;
      tim = i;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  read_and_prepare();
  shifts();
  cout << res << " " << tim << endl;
}
