#include <bits/stdc++.h>
using namespace std;
const int N = 1000000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, v[N], x, y, id;
long long bst;
vector<int> g;
multiset<int> st;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &v[i]);
    int d = v[i] - i;
    bst += abs(d);
    if (d > 0)
      ++x;
    else
      ++y;
    if (d > 0) st.insert(d);
  }
  long long cur = bst;
  for (int i = 1; i < n; ++i) {
    while (!st.empty() && *st.begin() - i < 0) {
      --x;
      ++y;
      st.erase(st.begin());
    }
    int d = v[n - i + 1], idx = n + 1 - i;
    int p = d - idx;
    int subt = min(max(0, p), n - idx), add = n - idx - subt;
    if (p - i >= 0) {
      st.erase(st.find(p));
      --x;
    } else
      --y;
    int off = d - 1;
    cur = cur + y - x - add + subt + off - abs(p);
    if (d - 1 > 0) {
      st.insert(d - 1 + i);
      ++x;
    } else
      ++y;
    if (cur < bst) {
      bst = cur;
      id = i;
    }
  }
  cout << bst << " " << id << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 1000000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, v[N], x, y, id;
long long bst;
vector<int> g;
multiset<int> st;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &v[i]);
    int d = v[i] - i;
    bst += abs(d);
    if (d > 0)
      ++x;
    else
      ++y;
    if (d > 0) st.insert(d);
  }
  long long cur = bst;
  for (int i = 1; i < n; ++i) {
    while (!st.empty() && *st.begin() - i < 0) {
      --x;
      ++y;
      st.erase(st.begin());
    }
    int d = v[n - i + 1], idx = n + 1 - i;
    int p = d - idx;
    int subt = min(max(0, p), n - idx), add = n - idx - subt;
    if (p - i >= 0) {
      st.erase(st.find(p));
      --x;
    } else
      --y;
    int off = d - 1;
    cur = cur + y - x - add + subt + off - abs(p);
    if (d - 1 > 0) {
      st.insert(d - 1 + i);
      ++x;
    } else
      ++y;
    if (cur < bst) {
      bst = cur;
      id = i;
    }
  }
  cout << bst << " " << id << endl;
}

```