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 maxn = 1e6 + 5;
long long p[maxn];
long long has[maxn << 1];
int main() {
  long long n;
  long long bigger = 0, smaller = 0, equ = 0, ans1 = 0, ans2 = 0;
  scanf("%I64d", &n);
  for (long long i = 1LL; i <= n; i++) {
    scanf("%I64d", &p[i]);
    if (p[i] > i) {
      bigger++;
      has[p[i] - i]++;
    } else if (p[i] == i) {
      equ++;
      has[0]++;
    } else
      smaller++;
    ans1 += abs(p[i] - i);
  }
  long long temp = ans1;
  for (long long last = n - 1LL, now = 1LL; last >= 1LL; last--, now++) {
    temp += (equ + smaller);
    temp -= bigger;
    smaller += equ;
    bigger -= has[now];
    if (p[last + 1] >= last + 1LL) has[p[last + 1] - last - 1]--;
    has[p[last + 1] - 1 + now]++;
    equ = has[now];
    if (p[last + 1] > 1LL) bigger++;
    smaller = n - equ - bigger;
    temp -= abs(p[last + 1] - n - 1LL);
    temp += abs(p[last + 1] - 1LL);
    if (temp < ans1) {
      ans1 = temp;
      ans2 = now;
    }
  }
  printf("%I64d %I64d\n", ans1, ans2);
  return 0;
}
