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 lint = long long int;
using pii = std::pair<int, int>;
using std::priority_queue;
using std::vector;
int num[1001000];
int N, adding, subing;
lint cur_ans;
priority_queue<pii, vector<pii>, std::greater<pii> > pq;
int main(void) {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int temp;
    scanf("%d", &temp);
    num[i] = temp;
    cur_ans += abs(temp - i);
    if (temp > i) {
      subing++;
      pq.push(pii(temp - i, i));
    } else {
      adding++;
    }
  }
  int cycling = 0;
  lint ans = cur_ans;
  for (int i = 1; i < N; i++) {
    cur_ans = cur_ans + adding - subing - 1;
    while (!pq.empty() && pq.top().first == i) {
      adding++;
      subing--;
      pq.pop();
    }
    cur_ans -= abs(num[N - i + 1] - N);
    cur_ans += abs(num[N - i + 1] - 1);
    adding--;
    if (num[N - i + 1] > 1) {
      subing++;
      pq.push(pii(num[N - i + 1] - 1 + i, N - i + 1));
    } else {
      adding++;
    }
    if (ans > cur_ans) {
      cycling = i;
      ans = cur_ans;
    }
  }
  printf("%lld %d", ans, cycling);
  return 0;
}
