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 n, a[1000008], use = 1;
pair<long long, long long> v;
int tr[4000007][5];
inline int myabs(int x) {
  if (x < 0) return -x;
  return x;
}
inline void insert(int x) {
  x += 1000002;
  int idx;
  int ptr = 0;
  for (int i = 21; i >= 0; i--) {
    idx = (1 << i & x) != 0;
    tr[ptr][idx + 2]++;
    if (!tr[ptr][idx]) tr[ptr][idx] = use++;
    ptr = tr[ptr][idx];
  }
}
inline void remove(int x) {
  x += 1000002;
  int idx;
  int ptr = 0;
  for (int i = 21; i >= 0; i--) {
    idx = (1 << i & x) != 0;
    tr[ptr][idx + 2]--;
    if (!tr[ptr][idx]) tr[ptr][idx] = use++;
    ptr = tr[ptr][idx];
  }
}
inline int g(int x) {
  x += 1000002;
  int cnt = 0, idx;
  int ptr = 0;
  for (int i = 21; i >= 0; i--) {
    idx = (1 << i & x) != 0;
    if (!idx) {
      cnt += tr[ptr][3];
    }
    ptr = tr[ptr][idx];
    if (ptr == 0) break;
  }
  return cnt;
}
int main() {
  scanf("%d", &n);
  int last = n;
  long long ans = 0;
  long long d;
  int k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int temp = a[i] - i;
    insert(temp);
    ans += myabs(temp);
  }
  d = ans;
  k = 0;
  for (int i = 0; i < n - 1; i++) {
    remove(a[last] - last);
    int temp = a[last] - last;
    int e = temp - i;
    ans -= myabs(temp);
    if (temp > 0) {
      if (e >= 0) {
        ans += i;
      } else {
        ans += temp;
        ans += e;
      }
    } else
      ans -= i;
    int gg = g(i);
    ans -= gg;
    ans += n - 1 - gg;
    insert((a[last] - 1) + (i + 1));
    ans += myabs(a[last] - 1);
    if (ans < d) {
      d = ans;
      k = i + 1;
    }
    last--;
  }
  printf("%lld %d\n", d, k);
  return 0;
}
