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;
using LL = long long;
constexpr LL LINF = 334ll << 53;
constexpr int INF = 15 << 26;
constexpr LL MOD = 1E9 + 7;
struct Problem {
  LL n;
  vector<LL> p;
  Problem(LL n) : n(n), p(n){};
  void solve() {
    LL ans = 0, minsum = LINF, sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> p[i];
      p[i]--;
    }
    multiset<LL> s;
    for (int i = 0; i < n; ++i) {
      if (p[i] - i > 0) s.insert(p[i] - i);
      sum += abs(p[i] - i);
    }
    minsum = sum;
    ;
    for (int i = 1; i < n; ++i) {
      LL cnt;
      if (p[n - i] - (n - 1) > 0)
        cnt = (int)s.size() - 1;
      else
        cnt = (int)s.size();
      sum += n - 1 - 2 * cnt;
      ;
      sum = sum - abs(p[n - i] - (n - 1)) + abs(p[n - i]);
      s.erase(i);
      if (p[n - i] > 0) s.insert(p[n - i] + i);
      ;
      ;
      if (sum < minsum) {
        minsum = sum;
        ans = i;
      }
    }
    cout << minsum << ' ' << ans << "\n";
  }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n = 0;
  cin >> n;
  Problem p(n);
  p.solve();
  return 0;
}
