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;
int Abs(int a) { return a < 0 ? -a : a; }
void add(vector<int> &BIT, int i, int val, int N) {
  while (i <= N) {
    BIT[i] += val;
    i += i & (-i);
  }
}
int query(vector<int> &BIT, int i) {
  int ans = 0;
  while (i > 0) {
    ans += BIT[i];
    i -= i & (-i);
  }
  return ans;
}
int main() {
  int N, n;
  int po;
  int tmp;
  long long sum = 0;
  long long ans;
  scanf("%d", &n);
  vector<int> BIT((n << 3) + 1), arr(n + 1), sav(n + 1);
  N = n << 3;
  po = n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    sum += Abs(arr[i] - i);
    add(BIT, arr[i] - i + po, 1, N);
    sav[arr[i]] = arr[i] - i + po;
  }
  ans = sum;
  tmp = 0;
  for (int i = 1, k = n; i < n; i++, k--) {
    add(BIT, sav[arr[k]], -1, N);
    int p1 = query(BIT, po);
    int p2 = query(BIT, N) - query(BIT, po);
    po++;
    add(BIT, arr[k] - 1 + po, 1, N);
    sum = sum + p1 - p2 + (arr[k] - 1) - Abs(arr[k] - n);
    if (ans > sum) {
      ans = sum;
      tmp = i;
    }
    sav[arr[k]] = po + arr[k] - 1;
  }
  printf("%lld %d\n", ans, tmp);
  return 0;
}
