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 int N = 1e6 + 5;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n;
int a[N];
long long ans1, ans2;
long long lft, rht;
int lsz, rsz;
int cnt[N << 1];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) cnt[a[i] - i + n]++;
  for (int i = 1; i <= n; i++)
    if (a[i] - i > 0) rsz++;
  for (int i = 1; i <= n; i++)
    if (a[i] - i <= 0) lsz++;
  for (int i = 1; i <= n; i++)
    if (a[i] - i <= 0) lft += i - a[i];
  for (int i = 1; i <= n; i++)
    if (a[i] - i > 0) rht += a[i] - i;
  ans1 = lft + rht, ans2 = 0;
  for (int i = 1; i <= n - 1; i++) {
    int id = n - i + 1;
    cnt[a[id] - id + n]--;
    lft -= n - a[id];
    lsz--;
    rht -= rsz;
    lft += lsz;
    rsz -= cnt[i + n];
    lsz += cnt[i + n];
    if (a[id] > 1) {
      rsz++;
      rht += a[id] - 1;
    } else
      lsz++;
    if (a[id] + i - 1 <= n) cnt[a[id] + i - 1 + n]++;
    if (lft + rht < ans1) ans1 = lft + rht, ans2 = i;
  }
  printf("%lld %lld\n", ans1, ans2);
  return 0;
}
