#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long n, a[1 << 21], b[1 << 21], lst;
pair<long long, long long> ans = {1e18, -1};
void add(long long l, long long r, long long val, long long pos) {
  if (l > r) return;
  if (r >= pos && l <= pos) {
    b[0] += val;
    b[r - pos + 1] -= val;
    b[n - pos + l] += val;
    b[n - pos + pos + 1] -= val;
  } else if (r >= pos && l >= pos) {
    b[l - pos] += val;
    b[r - pos + 1] -= val;
  } else {
    b[n - pos + l] += val;
    b[n - pos + r + 1] -= val;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    lst += abs(a[i] - i);
    add(2, a[i], -1, i);
    add(a[i] + 1, n, 1, i);
    add(1, 1, abs(1 - a[i]) - abs(n - a[i]), i);
  }
  for (long long i = 0; i < n; i++) {
    if (i) {
      b[i] += b[i - 1];
      lst += b[i];
    }
    if (lst < ans.first) ans = {lst, i};
  }
  return cout << ans.first << ' ' << ans.second, 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("O3")
using namespace std;
long long n, a[1 << 21], b[1 << 21], lst;
pair<long long, long long> ans = {1e18, -1};
void add(long long l, long long r, long long val, long long pos) {
  if (l > r) return;
  if (r >= pos && l <= pos) {
    b[0] += val;
    b[r - pos + 1] -= val;
    b[n - pos + l] += val;
    b[n - pos + pos + 1] -= val;
  } else if (r >= pos && l >= pos) {
    b[l - pos] += val;
    b[r - pos + 1] -= val;
  } else {
    b[n - pos + l] += val;
    b[n - pos + r + 1] -= val;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    lst += abs(a[i] - i);
    add(2, a[i], -1, i);
    add(a[i] + 1, n, 1, i);
    add(1, 1, abs(1 - a[i]) - abs(n - a[i]), i);
  }
  for (long long i = 0; i < n; i++) {
    if (i) {
      b[i] += b[i - 1];
      lst += b[i];
    }
    if (lst < ans.first) ans = {lst, i};
  }
  return cout << ans.first << ' ' << ans.second, 0;
}

```