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 double eps = 1e-8;
const long long INFF = 1e18;
const int INF = 1e9;
const int mod = 1e9 + 7;
const int MAX = 2e6 + 7;
int n, per[MAX];
int cnt[MAX << 1];
int main(void) {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); ++i) scanf("%d", per + i);
  int ans = 0;
  long long val = 0;
  for (int i = (1); i < (n + 1); ++i) val += abs(per[i] - i);
  for (int i = (1); i < (n + 1); ++i) cnt[per[i] - i + n]++;
  long long cur = val, num = 0;
  for (int i = (1); i < (n + 1); ++i)
    if (per[i] > i) num++;
  for (int i = (1); i < (n); ++i) {
    int bk = n + 1 - i;
    cnt[per[bk] - bk + n]--;
    if (per[bk] - bk + 1 - i > 0) num--;
    cur = cur - num + n - 1 - num;
    cur = cur - abs(per[bk] - bk + 1 - i) + abs(per[bk] - 1);
    num -= cnt[i + n];
    if (per[bk] > 1) num++;
    cnt[per[bk] - 1 + n + i]++;
    if (cur < val) {
      val = cur, ans = i;
    }
  }
  printf("%lld %d\n", val, ans);
  return 0;
}
