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;
using ii = pair<int, int>;
const int N = 1e6 + 666;
ll p[N], a[N], d[N], n;
void query(int l, int r, ll x, ll y) {
  if (l > r) return;
  a[l] += x;
  a[r + 1] -= x + (r - l) * y;
  d[l + 1] += y;
  d[r + 1] -= y;
}
void process() {
  for (int i = 0; i < n; i++) d[i] += d[i - 1];
  for (int i = 0; i < n; i++) a[i] += a[i - 1] + d[i];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) {
    vector<pair<int, ii>> r;
    if (i == p[i]) {
      r.push_back({1, ii(0, n - p[i])});
      r.push_back({-1, ii(p[i] - 1, 1)});
    } else if (i < p[i]) {
      r.push_back({-1, ii(p[i] - i, 1)});
      r.push_back({1, ii(0, n - p[i])});
      r.push_back({-1, ii(p[i] - 1, p[i] - i + 1)});
    } else if (i > p[i]) {
      r.push_back({1, ii(i - p[i], n - p[i])});
      r.push_back({-1, ii(p[i] - 1, 1)});
      r.push_back({1, ii(0, i - p[i] - 1)});
    }
    vector<pair<int, ii>> v;
    for (auto& x : r) {
      ii t = x.second;
      if (x.first == 1 && t.first > t.second) continue;
      if (x.first == -1 && t.first < t.second) continue;
      v.push_back(x);
    }
    int done = 0;
    for (auto& x : v) {
      ii t = x.second;
      int rsz =
          (x.first == 1 ? t.second - t.first + 1 : t.first - t.second + 1);
      query(done, done + rsz - 1, t.first, x.first);
      done += rsz;
    }
    assert(done == n);
  }
  process();
  ll mn = a[0], ans = 0;
  for (int i = 1; i < n; i++) {
    if (a[i] < mn) {
      mn = a[i];
      ans = i;
    }
  }
  cout << mn << ' ' << ans << '\n';
}
