#include <bits/stdc++.h>
using namespace std;
static const long long oo = 0x3f3f3f3f3f3f3f3fLL;
static const double eps = 1e-9;
template <class C, class I>
bool has(const C &con, const I &item) {
  return con.find(item) != con.end();
}
const long long maxN = 2 * 1024 * 1024;
long long p[maxN];
long long change[maxN];
int main() {
  ios::sync_with_stdio(false);
  long long N;
  cin >> N;
  for (long long i = (1); i < (N + 1); i++) {
    long long tmp;
    cin >> tmp;
    p[i] = tmp;
  }
  long long get_worse = 0;
  long long get_better = 0;
  long long devi = 0;
  for (long long i = (1); i < (N + 1); i++) {
    devi += abs(p[i] - i);
    if (i >= p[i])
      change[N - (i - p[i])]++;
    else
      change[p[i] - i]++;
    if (i < N) {
      if (i < p[i])
        get_better++;
      else
        get_worse++;
    }
  }
  pair<long long, long long> best{devi, 0};
  for (long long k = (1); k < (N); k++) {
    devi = devi + get_worse - get_better;
    devi -= abs(p[N + 1 - k] - N);
    devi += abs(p[N + 1 - k] - 1);
    get_worse += change[k] - 1;
    get_better -= change[k] - 1;
    best = min(best, pair<long long, long long>(devi, k));
  }
  cout << best.first << " " << best.second << endl;
}

### Prompt

Generate 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>
using namespace std;
static const long long oo = 0x3f3f3f3f3f3f3f3fLL;
static const double eps = 1e-9;
template <class C, class I>
bool has(const C &con, const I &item) {
  return con.find(item) != con.end();
}
const long long maxN = 2 * 1024 * 1024;
long long p[maxN];
long long change[maxN];
int main() {
  ios::sync_with_stdio(false);
  long long N;
  cin >> N;
  for (long long i = (1); i < (N + 1); i++) {
    long long tmp;
    cin >> tmp;
    p[i] = tmp;
  }
  long long get_worse = 0;
  long long get_better = 0;
  long long devi = 0;
  for (long long i = (1); i < (N + 1); i++) {
    devi += abs(p[i] - i);
    if (i >= p[i])
      change[N - (i - p[i])]++;
    else
      change[p[i] - i]++;
    if (i < N) {
      if (i < p[i])
        get_better++;
      else
        get_worse++;
    }
  }
  pair<long long, long long> best{devi, 0};
  for (long long k = (1); k < (N); k++) {
    devi = devi + get_worse - get_better;
    devi -= abs(p[N + 1 - k] - N);
    devi += abs(p[N + 1 - k] - 1);
    get_worse += change[k] - 1;
    get_better -= change[k] - 1;
    best = min(best, pair<long long, long long>(devi, k));
  }
  cout << best.first << " " << best.second << endl;
}

```