#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<long long> p(n);
  for (long long i = 0; i < n; i++) cin >> p[i];
  vector<long long> dif(n);
  long long sum = 0, pls = 0, mns = 0;
  for (long long i = 0; i < n; i++)
    dif[i] = p[i] - (i + 1), sum += abs(dif[i]), (dif[i] > 0 ? mns : pls)++;
  vector<long long> difs(n);
  difs[0] = sum;
  vector<long long> zeros(n);
  for (long long i = 0; i < n; i++) {
    long long d = (n + p[i] - i - 1) % n;
    zeros[d]++;
  }
  for (long long i = 1; i < n; i++) {
    sum += pls - mns;
    long long r = n - i;
    sum = sum - (n + 1 - p[r]) + (p[r] - 1);
    difs[i] = sum;
    pls--;
    mns++;
    pls += zeros[i];
    mns -= zeros[i];
  }
  long long mn = 1e18, mnidx;
  for (long long i = 0; i < n; i++)
    if (difs[i] < mn) mn = difs[i], mnidx = i;
  cout << mn << " " << mnidx << endl;
}

### Prompt

Please provide a CPP coded solution 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 main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<long long> p(n);
  for (long long i = 0; i < n; i++) cin >> p[i];
  vector<long long> dif(n);
  long long sum = 0, pls = 0, mns = 0;
  for (long long i = 0; i < n; i++)
    dif[i] = p[i] - (i + 1), sum += abs(dif[i]), (dif[i] > 0 ? mns : pls)++;
  vector<long long> difs(n);
  difs[0] = sum;
  vector<long long> zeros(n);
  for (long long i = 0; i < n; i++) {
    long long d = (n + p[i] - i - 1) % n;
    zeros[d]++;
  }
  for (long long i = 1; i < n; i++) {
    sum += pls - mns;
    long long r = n - i;
    sum = sum - (n + 1 - p[r]) + (p[r] - 1);
    difs[i] = sum;
    pls--;
    mns++;
    pls += zeros[i];
    mns -= zeros[i];
  }
  long long mn = 1e18, mnidx;
  for (long long i = 0; i < n; i++)
    if (difs[i] < mn) mn = difs[i], mnidx = i;
  cout << mn << " " << mnidx << endl;
}

```