#include <bits/stdc++.h>
using namespace std;
const int N = 1000002;
int n, p[N], sign[N];
long long curr;
vector<vector<int> > flips(N), add(N);
int move(int i, int d) {
  if (d < 1) return i;
  return (i + d) % n ? (i + d) % n : n;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n); ++i) {
    scanf("%d", &p[i]);
    curr += i - p[i];
    if (p[i] == n)
      flips[0].push_back(i), add[n - i + 1].push_back(i);
    else {
      if (i <= p[i]) flips[0].push_back(i);
      int j = p[i] + 1 <= n ? p[i] + 1 : 1;
      int k = i <= j ? j - i : n - i + j;
      if (k) flips[k].push_back(i);
      int l = i == 1 ? 0 : n - i + 1;
      if (l and l != k) flips[l].push_back(i);
    }
  }
  int ans = -1, pls = n;
  long long dev = 1000000000000000000ll;
  for (int k = (0); k <= (int)(n - 1); ++k) {
    int changedPlus = 0, changedMinus = 0;
    for (int i : flips[k]) {
      if (sign[i])
        curr = curr - (p[i] - move(i, k - 1)) + (move(i, k) - p[i]),
        ++changedMinus;
      else
        curr = curr - (move(i, k - 1) - p[i]) + (p[i] - move(i, k)),
        ++changedPlus;
      sign[i] ^= 1;
    }
    for (int i : add[k]) curr += p[i];
    if (k) curr = curr + (pls - changedPlus) - (n - pls - changedMinus);
    pls = pls + changedMinus - changedPlus;
    if (curr < dev) dev = curr, ans = k;
  }
  cout << dev << " " << ans;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 1000002;
int n, p[N], sign[N];
long long curr;
vector<vector<int> > flips(N), add(N);
int move(int i, int d) {
  if (d < 1) return i;
  return (i + d) % n ? (i + d) % n : n;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n); ++i) {
    scanf("%d", &p[i]);
    curr += i - p[i];
    if (p[i] == n)
      flips[0].push_back(i), add[n - i + 1].push_back(i);
    else {
      if (i <= p[i]) flips[0].push_back(i);
      int j = p[i] + 1 <= n ? p[i] + 1 : 1;
      int k = i <= j ? j - i : n - i + j;
      if (k) flips[k].push_back(i);
      int l = i == 1 ? 0 : n - i + 1;
      if (l and l != k) flips[l].push_back(i);
    }
  }
  int ans = -1, pls = n;
  long long dev = 1000000000000000000ll;
  for (int k = (0); k <= (int)(n - 1); ++k) {
    int changedPlus = 0, changedMinus = 0;
    for (int i : flips[k]) {
      if (sign[i])
        curr = curr - (p[i] - move(i, k - 1)) + (move(i, k) - p[i]),
        ++changedMinus;
      else
        curr = curr - (move(i, k - 1) - p[i]) + (p[i] - move(i, k)),
        ++changedPlus;
      sign[i] ^= 1;
    }
    for (int i : add[k]) curr += p[i];
    if (k) curr = curr + (pls - changedPlus) - (n - pls - changedMinus);
    pls = pls + changedMinus - changedPlus;
    if (curr < dev) dev = curr, ans = k;
  }
  cout << dev << " " << ans;
}

```