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 = 1e6 + 10;
int cnt[maxn];
int a[maxn], n;
int abs(int x) { return x > 0 ? x : -x; }
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n) {
    int add = 0, sub = 0, b = 0;
    long long tmp = 0, res;
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      tmp += abs(a[i] - i);
      cnt[(a[i] - i + n) % n]++;
      if (a[i] > i)
        sub++;
      else
        add++;
    }
    res = tmp;
    for (int i = 1; i < n; i++) {
      tmp += add - sub - 1;
      tmp += (a[n - i + 1] - 1) - (n - a[n - i + 1]);
      sub = sub - cnt[i] + 1;
      add = add + cnt[i] - 1;
      if (res > tmp) {
        res = tmp;
        b = i;
      }
    }
    cout << res << " " << b << endl;
  }
}
