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 a[1000005];
int mp[10000005];
int main() {
  int i, j, k, l, m, n, t, q, r;
  t = 1;
  while (t--) {
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    long long pos = 0;
    int np = 0;
    for (i = 1; i <= n; i++) {
      mp[a[i] - i + 3000000]++;
      if (a[i] - i > 0) {
        pos += (a[i] - i);
        np++;
      }
    }
    long long mi = pos;
    int id = 0;
    int sh = 1;
    for (i = n; i > 1; i--) {
      pos -= np;
      pos += (a[i] - 1);
      if (a[i] - 1 > 0) {
        np++;
      }
      np -= mp[(sh - 1) + 1 + 3000000];
      if (pos < mi) {
        mi = pos;
        id = sh;
      }
      mp[(a[i] - 1) + sh + 3000000]++;
      mp[(a[i] - n) + sh + 3000000]--;
      sh++;
    }
    cout << mi * 2 << " " << id << endl;
  }
}
