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 n, arr[1000010];
long long add[1000010][2];
inline int rd() {
  int a = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) (a *= 10) += c - '0', c = getchar();
  return a;
}
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) {
    arr[i] = rd();
    if (arr[i] >= i) {
      add[0][0] += arr[i] - i, add[0][1]--;
      add[arr[i] - i + 1][0] += 2, add[arr[i] - i + 1][1] += 2;
      add[n - i + 1][0] += arr[i] * 2 - n - 2, add[n - i + 1][1] -= 2;
    } else {
      add[0][0] += i - arr[i], add[0][1]++;
      add[n - i + 1][0] += arr[i] * 2 - n - 2, add[n - i + 1][1] -= 2;
      add[n - i + arr[i] + 1][0] += 2, add[n - i + arr[i] + 1][1] += 2;
    }
  }
  long long sum = 0, delta = 0, mans = 0x3f3f3f3f3f3f3f3fll;
  int mp = 0;
  for (int i = 0; i < n; i++) {
    sum += delta + add[i][0];
    if (sum < mans) mp = i, mans = sum;
    delta += add[i][1];
  }
  printf("%I64d %d\n", mans, mp);
  return 0;
}
