#include <bits/stdc++.h>
using namespace std;
int N;
int P[1010101];
signed long long dif[1010101];
signed long long sum[1010101];
signed long long V[1010101];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  scanf("%d", &N);
  for (i = 0; i < (N); i++) {
    scanf("%d", &P[i]), P[i]--;
    V[0] += abs(P[i] - i);
    if (P[i] == i) {
      dif[0]++;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N - i + 1]--;
      dif[N - i + 1]--;
    } else if (P[i] > i) {
      dif[0]--;
      dif[P[i] - i + 1]++;
      dif[P[i] - i + 1]++;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N - i + 1]--;
      dif[N - i + 1]--;
    } else {
      dif[0]++;
      dif[N - i + 1]--;
      dif[N - i + 1]--;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N + P[i] - i + 1]++;
      dif[N + P[i] - i + 1]++;
    }
  }
  signed long long shi = 0, ret = V[0];
  for (i = 1; i < N; i++) {
    dif[i] += dif[i - 1];
    V[i] += V[i - 1] + dif[i] + sum[i];
    if (V[i] < ret) shi = i, ret = V[i];
  }
  cout << ret << " " << shi << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
int P[1010101];
signed long long dif[1010101];
signed long long sum[1010101];
signed long long V[1010101];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  scanf("%d", &N);
  for (i = 0; i < (N); i++) {
    scanf("%d", &P[i]), P[i]--;
    V[0] += abs(P[i] - i);
    if (P[i] == i) {
      dif[0]++;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N - i + 1]--;
      dif[N - i + 1]--;
    } else if (P[i] > i) {
      dif[0]--;
      dif[P[i] - i + 1]++;
      dif[P[i] - i + 1]++;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N - i + 1]--;
      dif[N - i + 1]--;
    } else {
      dif[0]++;
      dif[N - i + 1]--;
      dif[N - i + 1]--;
      sum[N - i] -= N - P[i];
      sum[N - i] += P[i];
      dif[N + P[i] - i + 1]++;
      dif[N + P[i] - i + 1]++;
    }
  }
  signed long long shi = 0, ret = V[0];
  for (i = 1; i < N; i++) {
    dif[i] += dif[i - 1];
    V[i] += V[i - 1] + dif[i] + sum[i];
    if (V[i] < ret) shi = i, ret = V[i];
  }
  cout << ret << " " << shi << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}

```