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 inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long double eps = 1e-10;
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> P(n);
  for (int i = (int)(0); i < (int)(n); i++) cin >> P[i];
  vector<int> cnt(n);
  for (int i = (int)(0); i < (int)(n); i++) cnt[(P[i] - i - 1 + n) % n]++;
  long long tmp = 0;
  long long mn = longinf;
  int mni = 0;
  int h = 0, l = 0;
  for (int i = (int)(0); i < (int)(n); i++) {
    tmp += abs(P[i] - i - 1);
    (P[i] - i - 1 > 0 ? h : l)++;
  }
  for (int i = (int)(0); i < (int)(n); i++) {
    if (mn > tmp) {
      mn = tmp;
      mni = i;
    }
    tmp -= abs(P[n - 1 - i] - n);
    tmp += abs(P[n - 1 - i] - 1);
    tmp += (l - 1) - h;
    l += cnt[(i + 1) % n];
    h -= cnt[(i + 1) % n];
    l--;
    h++;
  }
  cout << mn << ' ' << mni << '\n';
  return 0;
}
