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 d[4000000];
int a[1100000];
void add1(int x, int l, int r, int ll, int rr, int v) {
  int mid;
  if (r < ll) return;
  if (rr < l) return;
  if ((ll <= l) && (r <= rr)) {
    d[x] += v;
    return;
  }
  mid = (l + r) / 2;
  add1(x + x + 1, l, mid, ll, rr, v);
  add1(x + x + 2, mid + 1, r, ll, rr, v);
}
long long get1(int x, int l, int r, int p) {
  long long ret;
  int mid;
  if ((l <= p) && (p <= r)) {
    if (l == r) return d[x];
    mid = (l + r) / 2;
    ret = 0;
    ret += get1(x + x + 1, l, mid, p);
    ret += get1(x + x + 2, mid + 1, r, p);
    return ret + d[x];
  } else
    return 0;
}
int main() {
  long long sum, bs, tmp;
  int i, j, n, bi;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  memset(d, 0, sizeof(d));
  for (i = 0; i < n; i++) {
    if (a[i] <= i) {
      if (n - 1 - i >= 1) {
        add1(0, 0, n - 1, 1, n - 1 - i, 1);
      }
    } else {
      if (a[i] - i >= 1) {
        add1(0, 0, n - 1, 1, a[i] - i, -1);
      }
      if (a[i] - i + 1 <= n - 1 - i) {
        add1(0, 0, n - 1, a[i] - i + 1, n - 1 - i, 1);
      }
    }
    if (n - i <= n - 1) add1(0, 0, n - 1, n - i, n - i, a[i] - (n - 1 - a[i]));
    if (a[i] < i) {
      if (n - i + a[i] >= n - i + 1)
        add1(0, 0, n - 1, n - i + 1, n - i + a[i], -1);
      if (n - 1 >= n - i + a[i] + 1)
        add1(0, 0, n - 1, n - i + a[i] + 1, n - 1, 1);
    } else {
      if (n - 1 >= n - i + 1) add1(0, 0, n - 1, n - i + 1, n - 1, -1);
    }
  }
  sum = 0;
  for (i = 0; i < n; i++) {
    sum += abs(a[i] - i);
  }
  bs = sum;
  bi = 0;
  for (i = 1; i < n; i++) {
    tmp = get1(0, 0, n - 1, i);
    sum += tmp;
    if (sum < bs) {
      bs = sum;
      bi = i;
    }
  }
  cout << bs << " " << bi << endl;
  return 0;
}
