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;
template <typename T>
inline void read(T &FF) {
  int RR = 1;
  FF = 0;
  char CH = getchar();
  for (; !isdigit(CH); CH = getchar())
    if (CH == '-') RR = -RR;
  for (; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
  FF *= RR;
}
inline void file(string str) {
  freopen((str + ".in").c_str(), "r", stdin);
  freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 2e6 + 10;
int n, seq[N], tax[N], li, ri;
int main() {
  read(n);
  long long res = 0, ans = 0;
  int id = 0;
  for (int i = 1; i <= n; i++) {
    read(seq[i]);
    if (i < n) {
      if (seq[i] <= i)
        ri++;
      else
        tax[seq[i] - i]++, li++;
    }
    res += abs(seq[i] - i);
  }
  ans = res;
  for (int i = 1; i <= n - 1; i++) {
    res = res - li + ri;
    res = res + 2 * seq[n - i + 1] - n - 1;
    li -= tax[i], ri += tax[i];
    if (seq[n - i + 1] != 1) ri--, li++;
    tax[i + seq[n - i + 1] - 1]++;
    if (res < ans) ans = res, id = i;
  }
  cout << ans << " " << id << endl;
  return 0;
}
