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;
inline int Read() {
  char c = getchar();
  int num = 0;
  while ('0' > c || c > '9') c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  return (num);
}
long long mn, s0, s1;
int nm[1000010 * 2], n, a[1000010], ans, n0, n1;
inline int Abs(int x) {
  if (x >= 0) return (x);
  return (-x);
}
int main() {
  n = Read();
  for (int i = 1; i <= n; i++) a[i] = Read();
  for (int i = 1; i <= n; i++) {
    if (a[i] - i >= 0)
      nm[a[i] - i]++, s0 += a[i] - i, n0++;
    else
      s1 += i - a[i], n1++;
    mn += Abs(a[i] - i);
  }
  ans = 0;
  for (int i = 1; i < n; i++) {
    n1 += nm[i - 1], s1 += n1;
    n0 -= nm[i - 1], s0 -= n0;
    nm[i - 1] = 0;
    s1 -= (n + 1 - a[n - i + 1]), n1--;
    s0 += a[n - i + 1] - 1, n0++, nm[a[n - i + 1] - 1 + i]++;
    if (mn > s0 + s1) mn = s0 + s1, ans = i;
  }
  printf("%I64d %d\n", mn, ans);
}
