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;
const long long oo = 0x3f3f3f3f3f3f3f3fll;
const int LGN = 25;
const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
int n;
long long delta[N];
void add(int l, int r, int v) {
  if (r < l || n <= l) return;
  delta[l] += v;
  delta[r + 1] -= v;
}
int main() {
  std::ios::sync_with_stdio(false);
  scanf("%d", &(n));
  static int p[N];
  long long d = 0;
  for (long long i = 1, __R = n; i <= __R; i++) {
    scanf("%d", &(p[i]));
    d += abs(i - p[i]);
    if (p[i] > i) {
      add(1, p[i] - i, -1);
      add(p[i] - i + 1, n - i, +1);
      add(n - i + 1, n - i + 1, -(n - p[i]) + (p[i] - 1));
      add(n - i + 2, n - 1, -1);
    } else if (p[i] == i) {
      add(1, n - i, +1);
      add(n - i + 1, n - i + 1, -(n - p[i]) + (p[i] - 1));
      add(n - i + 2, n - 1, -1);
    } else {
      add(1, n - i, +1);
      add(n - i + 1, n - i + 1, -(n - p[i]) + (p[i] - 1));
      add(n - i + 2, n - i + p[i], -1);
      add(n - i + p[i] + 1, n - 1, +1);
    }
  }
  long long mind = d;
  int mink = 0;
  for (long long k = 1, __R = n - 1; k <= __R; k++) {
    delta[k] += delta[k - 1];
  }
  for (long long k = 1, __R = n - 1; k <= __R; k++) {
    d += delta[k];
    if (d < mind) {
      mind = d;
      mink = k;
    }
  }
  printf("%lld %d\n", mind, mink);
  return 0;
}
