#include <bits/stdc++.h>
using namespace std;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 4e6 + 5;
int n;
int p[2 * N];
int cnt[2 * N];
long long sum[2 * N];
void add(int pos, int val) {
  for (int cur = pos; cur < 2 * N; cur |= (cur + 1)) {
    cnt[cur] += val;
    sum[cur] += 1ll * val * (pos - 2 * n);
  }
}
pair<int, long long> get(int l, int r) {
  int cn = 0;
  long long sm = 0;
  for (int cur = r; cur >= 0; cur = (cur & (cur + 1)) - 1) {
    cn += cnt[cur];
    sm += sum[cur];
  }
  for (int cur = l - 1; cur >= 0; cur = (cur & (cur + 1)) - 1) {
    cn -= cnt[cur];
    sm -= sum[cur];
  }
  return make_pair(cn, sm);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    p[i + n] = p[i];
  }
  for (int i = 1; i <= 2 * n; i++) {
    p[i] = p[i] - i;
  }
  for (int i = 1; i <= n; i++) {
    add(p[i] + 2 * n, 1);
  }
  long long best_val = inf64;
  int best_id = -1;
  pair<int, long long> tmp;
  long long tmp_val;
  for (int ql, qr, de = 0, l = 1, r = n; r <= 2 * n; l++, r++, de++) {
    tmp_val = 0;
    if (0 < 2 * n - de) {
      tmp = get(0, 2 * n - de - 1);
      tmp_val += (-tmp.second);
      tmp_val += 1ll * (-de) * tmp.first;
    }
    if (-de + 2 * n <= 3 * n) {
      tmp = get(2 * n - de, 3 * n);
      tmp_val += tmp.second;
      tmp_val += 1ll * de * tmp.first;
    }
    if (best_val > tmp_val) {
      best_val = tmp_val;
      best_id = ((-de) % n + n) % n;
    }
    add(p[l] + 2 * n, -1);
    if (r < 2 * n) {
      add(p[r + 1] + 2 * n, 1);
    }
  }
  cout << best_val << " " << best_id << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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 const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 4e6 + 5;
int n;
int p[2 * N];
int cnt[2 * N];
long long sum[2 * N];
void add(int pos, int val) {
  for (int cur = pos; cur < 2 * N; cur |= (cur + 1)) {
    cnt[cur] += val;
    sum[cur] += 1ll * val * (pos - 2 * n);
  }
}
pair<int, long long> get(int l, int r) {
  int cn = 0;
  long long sm = 0;
  for (int cur = r; cur >= 0; cur = (cur & (cur + 1)) - 1) {
    cn += cnt[cur];
    sm += sum[cur];
  }
  for (int cur = l - 1; cur >= 0; cur = (cur & (cur + 1)) - 1) {
    cn -= cnt[cur];
    sm -= sum[cur];
  }
  return make_pair(cn, sm);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    p[i + n] = p[i];
  }
  for (int i = 1; i <= 2 * n; i++) {
    p[i] = p[i] - i;
  }
  for (int i = 1; i <= n; i++) {
    add(p[i] + 2 * n, 1);
  }
  long long best_val = inf64;
  int best_id = -1;
  pair<int, long long> tmp;
  long long tmp_val;
  for (int ql, qr, de = 0, l = 1, r = n; r <= 2 * n; l++, r++, de++) {
    tmp_val = 0;
    if (0 < 2 * n - de) {
      tmp = get(0, 2 * n - de - 1);
      tmp_val += (-tmp.second);
      tmp_val += 1ll * (-de) * tmp.first;
    }
    if (-de + 2 * n <= 3 * n) {
      tmp = get(2 * n - de, 3 * n);
      tmp_val += tmp.second;
      tmp_val += 1ll * de * tmp.first;
    }
    if (best_val > tmp_val) {
      best_val = tmp_val;
      best_id = ((-de) % n + n) % n;
    }
    add(p[l] + 2 * n, -1);
    if (r < 2 * n) {
      add(p[r + 1] + 2 * n, 1);
    }
  }
  cout << best_val << " " << best_id << "\n";
  return 0;
}

```