#include <bits/stdc++.h>
using namespace std;
int p[1000005];
int N, nr1, nr2;
long long A[1000005];
long long B[1000005];
long long init;
long long eval(long long perm) {
  long long rez = 0;
  for (long long i = 1; i <= N; i++) {
    int ind = i + perm;
    ind = (ind - 1) % N + 1;
    rez += abs(ind - p[i]);
  }
  return rez;
}
int cand(int i, int pos) {
  if (i <= pos) return pos - i;
  return N - i + pos;
}
void addprog(int st, int dr, int p, int q) {
  A[st] += p;
  A[dr + 1] -= p;
  A[st] -= q * st;
  A[dr + 1] += q * st;
  B[st] += q;
  B[dr + 1] -= q;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> p[i];
    if (p[i] >= i) {
      addprog(0, p[i] - i, p[i] - i, -1);
      addprog(p[i] - i + 1, N - i, 1, 1);
      addprog(N - i + 1, N - 1, p[i] - 1, -1);
    } else {
      addprog(0, N - i, i - p[i], 1);
      addprog(N - i + 1, N - i + p[i], p[i] - 1, -1);
      addprog(N - i + p[i] + 1, N - 1, 1, 1);
    }
  }
  for (int i = 1; i <= N; i++) {
    A[i] += A[i - 1];
    B[i] += B[i - 1];
  }
  long long rez = 1LL << 60, poz;
  for (int i = 0; i < N; i++) {
    if (rez > A[i] + B[i] * i) {
      rez = A[i] + B[i] * i;
      poz = i;
    }
  }
  cout << rez << " " << poz;
  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;
int p[1000005];
int N, nr1, nr2;
long long A[1000005];
long long B[1000005];
long long init;
long long eval(long long perm) {
  long long rez = 0;
  for (long long i = 1; i <= N; i++) {
    int ind = i + perm;
    ind = (ind - 1) % N + 1;
    rez += abs(ind - p[i]);
  }
  return rez;
}
int cand(int i, int pos) {
  if (i <= pos) return pos - i;
  return N - i + pos;
}
void addprog(int st, int dr, int p, int q) {
  A[st] += p;
  A[dr + 1] -= p;
  A[st] -= q * st;
  A[dr + 1] += q * st;
  B[st] += q;
  B[dr + 1] -= q;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> p[i];
    if (p[i] >= i) {
      addprog(0, p[i] - i, p[i] - i, -1);
      addprog(p[i] - i + 1, N - i, 1, 1);
      addprog(N - i + 1, N - 1, p[i] - 1, -1);
    } else {
      addprog(0, N - i, i - p[i], 1);
      addprog(N - i + 1, N - i + p[i], p[i] - 1, -1);
      addprog(N - i + p[i] + 1, N - 1, 1, 1);
    }
  }
  for (int i = 1; i <= N; i++) {
    A[i] += A[i - 1];
    B[i] += B[i - 1];
  }
  long long rez = 1LL << 60, poz;
  for (int i = 0; i < N; i++) {
    if (rez > A[i] + B[i] * i) {
      rez = A[i] + B[i] * i;
      poz = i;
    }
  }
  cout << rez << " " << poz;
  return 0;
}

```