#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 1;
int p[MAXN];
struct Event {
  int x;
  long long val;
  int k;
  int num;
  Event(int x = -1, long long val = 0, int k = 0, int num = -1)
      : x(x), val(val), k(k), num(num) {}
};
bool compare(Event a, Event b) { return a.x < b.x; }
vector<Event> v;
Event f[MAXN];
long long get_val(Event ev, int x) {
  return ev.val + (long long)(x - ev.x) * (long long)ev.k;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (p[i] > i) {
      v.push_back(Event(0, p[i] - i, -1, i));
      v.push_back(Event(p[i] - i, 0, 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
    } else if (p[i] < i) {
      v.push_back(Event(0, i - p[i], 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
      v.push_back(Event(n - i + p[i], 0, 1, i));
    } else if (i == p[i]) {
      v.push_back(Event(0, 0, 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
    }
  }
  stable_sort(v.begin(), v.end(), compare);
  int last = 0;
  long long curans = 0;
  int k = 0;
  long long ans = 1e18;
  int ansp = -1;
  fill(f, f + MAXN, Event(0, 0, 0, 0));
  for (int i = 0; i < n; i++) {
    while (last < (int)v.size() && v[last].x <= i) {
      curans -= get_val(f[v[last].num], i);
      k -= f[v[last].num].k;
      f[v[last].num] = v[last];
      curans += v[last].val;
      k += v[last].k;
      last++;
    }
    if (curans < ans) ans = curans, ansp = i;
    curans += (long long)k;
  }
  cout << ans << " " << ansp << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 1;
int p[MAXN];
struct Event {
  int x;
  long long val;
  int k;
  int num;
  Event(int x = -1, long long val = 0, int k = 0, int num = -1)
      : x(x), val(val), k(k), num(num) {}
};
bool compare(Event a, Event b) { return a.x < b.x; }
vector<Event> v;
Event f[MAXN];
long long get_val(Event ev, int x) {
  return ev.val + (long long)(x - ev.x) * (long long)ev.k;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (p[i] > i) {
      v.push_back(Event(0, p[i] - i, -1, i));
      v.push_back(Event(p[i] - i, 0, 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
    } else if (p[i] < i) {
      v.push_back(Event(0, i - p[i], 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
      v.push_back(Event(n - i + p[i], 0, 1, i));
    } else if (i == p[i]) {
      v.push_back(Event(0, 0, 1, i));
      v.push_back(Event(n - i, p[i], -1, i));
    }
  }
  stable_sort(v.begin(), v.end(), compare);
  int last = 0;
  long long curans = 0;
  int k = 0;
  long long ans = 1e18;
  int ansp = -1;
  fill(f, f + MAXN, Event(0, 0, 0, 0));
  for (int i = 0; i < n; i++) {
    while (last < (int)v.size() && v[last].x <= i) {
      curans -= get_val(f[v[last].num], i);
      k -= f[v[last].num].k;
      f[v[last].num] = v[last];
      curans += v[last].val;
      k += v[last].k;
      last++;
    }
    if (curans < ans) ans = curans, ansp = i;
    curans += (long long)k;
  }
  cout << ans << " " << ansp << endl;
  return 0;
}

```