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;
void addmod(int &a, long long b) {
  a = (a + b);
  if (a >= 1000000007) a -= 1000000007;
}
void mulmod(int &a, long long b) { a = (a * b) % 1000000007; }
template <class T>
bool domin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool domax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gll() {
  long long a;
  scanf("%lld", &a);
  return a;
}
int main() {
  int n = gi(), ix = 0;
  vector<int> a(n + 1);
  long long ans = 0, nr = 0, ni = 0, mans = LLONG_MAX;
  vector<int> sr(n + 5, 0);
  for (auto i = 1; i <= n; i++) {
    a[i] = gi();
    ans += abs(i - a[i]);
    if (a[i] <= i) {
      ni++;
    } else {
      nr++;
      sr[a[i] - i]++;
    }
  }
  mans = ans;
  for (int i = n, j = 1; i > 1; i--, j++) {
    ni--;
    long long tans = ans - abs(a[i] - n);
    tans += a[i] - 1;
    tans += (ni - nr);
    if (mans > tans) {
      mans = tans;
      ix = j;
    }
    ans = tans;
    nr++;
    int k = j + a[i] - 1;
    if (k <= n) sr[k]++;
    nr -= sr[j];
    ni += sr[j];
  }
  cout << mans << " " << ix << endl;
  return 0;
}
