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 = 1000000 + 10;
long long a[MAXN];
long long b[MAXN];
void add(long long k, long long val, int start, int end) {
  if (start > end) return;
  a[start] += k;
  a[end + 1] -= k;
  long long bias = val - start * k;
  b[start] += bias;
  b[end + 1] -= bias;
}
int main() {
  int n, num;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &num);
    if (num >= i) {
      add(-1, num - i, 0, num - i);
      add(1, 1, num - i + 1, n - i);
      add(-1, num - 1, n - i + 1, n - 1);
    } else {
      add(1, i - num, 0, n - i);
      add(-1, num - 1, n - i + 1, n + num - i);
      add(1, 1, n + num - i + 1, n - 1);
    }
  }
  int pos = 0;
  long long val = b[0];
  for (int i = 1; i < n; i++) {
    a[i] += a[i - 1];
    b[i] += b[i - 1];
    if (a[i] * i + b[i] < val) {
      val = a[i] * i + b[i];
      pos = i;
    }
  }
  cout << val << ' ' << pos << endl;
  return 0;
}
