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;
int main(void) {
  cin >> n;
  vector<int> p(n);
  for (int i = (0); i < (n); i++) scanf("%d", &p[i]), p[i]--;
  vector<long long> t(n + 1);
  long long ch = 0;
  long long d = 0;
  for (int i = (0); i < (n); i++) {
    t[(p[i] - i + n) % n]++;
    if (p[i] > i)
      ch--;
    else
      ch++;
    d += abs(p[i] - i);
  }
  long long mk = 0;
  long long m = 1e18;
  t[n] = t[0];
  t[0] = 1;
  long long j = ch, l = d;
  for (int k = (0); k < (n + 1); k++) {
    if (k) d += ch - abs(n - 1 - p[n - k]) + abs(p[n - k]) - 1;
    if (d < m) m = d, mk = k;
    ch += 2 * t[k] - 2;
    if (k == n) assert(j == ch && l == d);
  }
  printf("%lld %lld\n", m, mk);
}
