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 read() {
  int x = 0, icon = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') icon = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * icon;
}
int _abs(int x) { return x < 0 ? -x : x; }
long long _min(long long a, long long b) { return a < b ? a : b; }
const int N = 2000000 + 233;
int a[N], num_s = 0, num_b = 0, c[N];
int ss[N], bb[N], n;
long long ans = 0, tmp, last;
int main() {
  n = read();
  for (int i = 0; i < n; i++) {
    a[i] = read();
    c[i] = a[i] - i - 1;
    if (c[i] > 0) {
      num_b++;
      bb[c[i]]++;
    }
    if (c[i] <= 0) {
      num_s++;
      ss[-c[i]]++;
    }
    ans += _abs(c[i]);
  }
  last = ans;
  int u = 0;
  for (int i = 1; i < n; i++) {
    tmp = last;
    tmp += num_s;
    tmp -= num_b;
    num_b -= bb[i];
    num_s += bb[i];
    c[n - i] -= i - 1;
    if (c[n - i] <= 0) {
      if (c[n - i] + n - 1 > 0) {
        int dd = c[n - i] + n - 1 - (_abs(c[n - i]) + 1);
        c[n - i] += n - 1;
        tmp += dd;
        num_b++;
        num_s--;
        bb[c[n - i] + i]++;
      } else
        tmp -= n;
    } else
      tmp += n - 2;
    if (ans > tmp) ans = tmp, u = i;
    last = tmp;
  }
  cout << ans << " " << u << endl;
  return 0;
}
