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;
long long o = 1000006;
long long m[4000008];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long p[n + 1];
  long long a, b = 0, c = 0, d = 0, i, j, k, x, y, y1, e, f, g, h;
  c = d = 0;
  for (i = 1; i <= n; i++) {
    cin >> p[i];
    a = p[i] - i;
    b += fabs(a);
    if (a <= 0)
      c++;
    else
      d++;
    if (m[a + o])
      m[a + o]++;
    else
      m[a + o] = 1;
  }
  x = 0;
  e = n;
  y1 = b;
  for (i = 1; i < n; i++) {
    f = p[e] - e;
    m[f + o]--;
    g = f - (i - 1);
    f += n;
    if (m[f + o])
      m[f + o]++;
    else
      m[f + o] = 1;
    c--;
    y = y1 + c - d;
    y -= fabs(g);
    g += n - 1;
    d++;
    y += fabs(g);
    if (y < b) {
      b = y;
      x = i;
    }
    if (m[i + o]) {
      c += m[i + o];
      d -= m[i + o];
    }
    y1 = y;
    e--;
  }
  cout << b << " " << x;
  return 0;
}
