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 = 6e6 + 1;
const int INF = 1e9 + 9;
const int B = 1e9 + 7;
int n;
int a[N];
int mp[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  scanf("%d", &n);
  int shift = 2 * n;
  long long s = 0;
  int l = 0, r = 0;
  for (int i = int(1); i <= int(n); ++i) {
    scanf("%d", &a[i]);
    ++mp[a[i] - i + shift];
    s += abs(a[i] - i);
    if (a[i] < i) ++l;
    if (a[i] > i) ++r;
  }
  long long ans = s, at = 0;
  int lst = n;
  for (int i = int(1); i <= int(n - 1); ++i) {
    s -= abs(a[lst] - n);
    --mp[a[lst] - lst + shift];
    if (a[lst] - lst > i - 1) --r;
    if (a[lst] - lst < i - 1) --l;
    l += mp[i - 1 + shift];
    s += +l - r;
    r -= mp[i + shift];
    s += abs(a[lst] - 1);
    ++mp[a[lst] - (1 - i) + shift];
    if (a[lst] - (1 - i) > i) ++r;
    if (a[lst] - (1 - i) < i) ++l;
    --lst;
    if (s < ans) ans = s, at = i;
  }
  cout << ans << " " << at << "\n";
  return 0;
}
