#include <bits/stdc++.h>
using namespace std;
int n;
const int maxn = 1000010;
int a[maxn];
struct yjq {
  int p, t;
  bool operator<(const yjq &an) const {
    if (t != an.t)
      return t > an.t;
    else
      return p < an.p;
  }
} tmp;
priority_queue<yjq> st;
int add;
int decr;
long long ans;
long long res;
inline bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    ans += max(a[i] - i, i - a[i]);
    if (a[i] <= i)
      add++;
    else {
      decr++;
      tmp.p = a[i];
      tmp.t = a[i] - i;
      st.push(tmp);
    }
  }
  res = ans;
  int t = 0;
  int num = 0;
  for (int i = n; i > 1; i--) {
    while ((!st.empty()) && st.top().t == t) {
      decr--;
      add++;
      st.pop();
    }
    tmp.t = a[i] + t;
    tmp.p = a[i];
    st.push(tmp);
    add--;
    decr++;
    ans += a[i] - 1 - (n - a[i]) + add - decr + 1;
    t++;
    if (ans < res) {
      res = ans;
      num = t;
    }
  }
  cout << res << " " << num << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
int n;
const int maxn = 1000010;
int a[maxn];
struct yjq {
  int p, t;
  bool operator<(const yjq &an) const {
    if (t != an.t)
      return t > an.t;
    else
      return p < an.p;
  }
} tmp;
priority_queue<yjq> st;
int add;
int decr;
long long ans;
long long res;
inline bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    ans += max(a[i] - i, i - a[i]);
    if (a[i] <= i)
      add++;
    else {
      decr++;
      tmp.p = a[i];
      tmp.t = a[i] - i;
      st.push(tmp);
    }
  }
  res = ans;
  int t = 0;
  int num = 0;
  for (int i = n; i > 1; i--) {
    while ((!st.empty()) && st.top().t == t) {
      decr--;
      add++;
      st.pop();
    }
    tmp.t = a[i] + t;
    tmp.p = a[i];
    st.push(tmp);
    add--;
    decr++;
    ans += a[i] - 1 - (n - a[i]) + add - decr + 1;
    t++;
    if (ans < res) {
      res = ans;
      num = t;
    }
  }
  cout << res << " " << num << endl;
  return 0;
}

```