#include <bits/stdc++.h>
const long long N = 1e6 + 100;
long long n;
long long p[N];
void init() {
  std::cin >> n;
  for (long long i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    p[i] = t;
  }
}
long long tag(long long x) { return x > 0 ? 1 : 0; }
void work() {
  static long long cnt[2][N << 1];
  long long sum[2] = {0, 0};
  long long size[2] = {0, 0};
  std::pair<long long, long long> answer;
  std::memset(cnt, 0, sizeof(cnt));
  for (long long i = 1; i <= n; i++) {
    cnt[tag(p[i] - i)][std::abs(p[i] - i)]++;
    size[tag(p[i] - i)]++;
    sum[tag(p[i] - i)] += std::abs(p[i] - i);
  }
  answer = {sum[0] + sum[1], 0};
  for (long long i = 1; i < n; i++) {
    sum[0] += size[0];
    sum[1] -= size[1];
    size[0] += cnt[1][i];
    size[1] -= cnt[1][i];
    long long oldValue = p[n - i + 1] - (n + 1);
    long long newValue = p[n - i + 1] - 1;
    sum[tag(oldValue)] -= std::abs(oldValue);
    sum[tag(newValue)] += std::abs(newValue);
    size[tag(oldValue)]--;
    size[tag(newValue)]++;
    cnt[tag(p[n - i + 1] - (n - i + 1))]
       [std::abs(p[n - i + 1] - (n - i + 1))]--;
    cnt[tag(p[n - i + 1] - (n - i + 1) + n)]
       [std::abs(p[n - i + 1] - (n - i + 1) + n)]++;
    answer = std::min(answer, {sum[0] + sum[1], i});
  }
  std::cout << answer.first << " " << answer.second << std::endl;
}
int main() {
  init();
  work();
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const long long N = 1e6 + 100;
long long n;
long long p[N];
void init() {
  std::cin >> n;
  for (long long i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    p[i] = t;
  }
}
long long tag(long long x) { return x > 0 ? 1 : 0; }
void work() {
  static long long cnt[2][N << 1];
  long long sum[2] = {0, 0};
  long long size[2] = {0, 0};
  std::pair<long long, long long> answer;
  std::memset(cnt, 0, sizeof(cnt));
  for (long long i = 1; i <= n; i++) {
    cnt[tag(p[i] - i)][std::abs(p[i] - i)]++;
    size[tag(p[i] - i)]++;
    sum[tag(p[i] - i)] += std::abs(p[i] - i);
  }
  answer = {sum[0] + sum[1], 0};
  for (long long i = 1; i < n; i++) {
    sum[0] += size[0];
    sum[1] -= size[1];
    size[0] += cnt[1][i];
    size[1] -= cnt[1][i];
    long long oldValue = p[n - i + 1] - (n + 1);
    long long newValue = p[n - i + 1] - 1;
    sum[tag(oldValue)] -= std::abs(oldValue);
    sum[tag(newValue)] += std::abs(newValue);
    size[tag(oldValue)]--;
    size[tag(newValue)]++;
    cnt[tag(p[n - i + 1] - (n - i + 1))]
       [std::abs(p[n - i + 1] - (n - i + 1))]--;
    cnt[tag(p[n - i + 1] - (n - i + 1) + n)]
       [std::abs(p[n - i + 1] - (n - i + 1) + n)]++;
    answer = std::min(answer, {sum[0] + sum[1], i});
  }
  std::cout << answer.first << " " << answer.second << std::endl;
}
int main() {
  init();
  work();
  return 0;
}

```