#include <bits/stdc++.h>
using namespace std;
vector<long long> p;
long long bst = 1e18;
long long bstloc = 1e9;
void adb(long long sc, long long lc) {
  if (sc < bst) {
    bst = sc;
    bstloc = lc;
  }
}
vector<long long> df;
long long tsm = 0;
long long tlg = 0;
long long tdev = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  scanf("%lld", &n);
  p.resize(n);
  df.resize(n);
  for (long long i = 0; i < (n); ++i) {
    scanf("%lld", &(p[i]));
    --p[i];
    if (p[i] > i) {
      ++tlg;
      tdev += p[i] - i;
      ++df[p[i] - i];
    } else {
      ++tsm;
      tdev += i - p[i];
      if (i > p[i]) ++df[p[i] - i + n];
    }
  }
  adb(tdev, 0);
  for (long long i = 1; i < n; ++i) {
    tdev -= (n - 1 - p[n - i]);
    tdev += p[n - i];
    --tsm;
    tdev -= tlg;
    tdev += tsm;
    ++tlg;
    tlg -= df[i];
    tsm += df[i];
    adb(tdev, i);
  }
  printf("%lld %lld\n", bst, bstloc);
  return 0;
}

### 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;
vector<long long> p;
long long bst = 1e18;
long long bstloc = 1e9;
void adb(long long sc, long long lc) {
  if (sc < bst) {
    bst = sc;
    bstloc = lc;
  }
}
vector<long long> df;
long long tsm = 0;
long long tlg = 0;
long long tdev = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  scanf("%lld", &n);
  p.resize(n);
  df.resize(n);
  for (long long i = 0; i < (n); ++i) {
    scanf("%lld", &(p[i]));
    --p[i];
    if (p[i] > i) {
      ++tlg;
      tdev += p[i] - i;
      ++df[p[i] - i];
    } else {
      ++tsm;
      tdev += i - p[i];
      if (i > p[i]) ++df[p[i] - i + n];
    }
  }
  adb(tdev, 0);
  for (long long i = 1; i < n; ++i) {
    tdev -= (n - 1 - p[n - i]);
    tdev += p[n - i];
    --tsm;
    tdev -= tlg;
    tdev += tsm;
    ++tlg;
    tlg -= df[i];
    tsm += df[i];
    adb(tdev, i);
  }
  printf("%lld %lld\n", bst, bstloc);
  return 0;
}

```