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() {
  int x = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n, k[1000009], g[1000009], A, B, ans2;
long long now, ans;
int main() {
  n = read();
  for (int i = 0; i <= n - 1; i++) k[i] = read() - 1, g[(k[i] - i + n) % n]++;
  for (int i = 0; i <= n - 1; i++)
    if (k[i] - i >= 0)
      A++, now += k[i] - i;
    else
      B++, now += i - k[i];
  ans = now, ans2 = 0;
  for (int i = 0; i <= n - 1; i++) {
    if (now < ans) ans = now, ans2 = i;
    now -= (n - 1) - k[n - 1 - i];
    B--;
    A -= g[i];
    B += g[i];
    now -= A;
    now += B;
    now += k[n - 1 - i];
    A++;
  }
  printf("%lld %d\n", ans, ans2);
  return 0;
}
