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 MAX = 1 << 20;
int perm[MAX], change[MAX];
void solve(int test_num) {
  (void)test_num;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", perm + i);
    --perm[i];
  }
  int leq = 0;
  long long sum = 0;
  memset(change, 0, sizeof(change));
  for (int i = 0; i < n; ++i) {
    sum += abs(perm[i] - i);
    if (perm[i] > i)
      ++change[perm[i] - i];
    else {
      ++leq;
      ++change[n - (i - perm[i])];
    }
  }
  pair<long long, int> res(sum, 0);
  for (int k = 1; k < n; ++k) {
    sum += leq - (n - leq);
    leq += change[k];
    sum -= abs(perm[n - k] - n);
    sum += abs(perm[n - k]);
    --leq;
    res = min(res, {sum, k});
  }
  cout << res.first << " " << res.second << endl;
}
void init() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  srand(1223);
  init();
  int tests = 1;
  for (int test = 1; test <= tests; ++test) {
    solve(test);
  }
  return 0;
}
