#include <bits/stdc++.h>
using namespace std;
long long gcd(long long n1, long long n2) {
  if (n2 == 0) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  if (exponent < 0) exponent += 1000000007LL - 1;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007LL;
    base = (base * base) % 1000000007LL;
    exponent /= 2;
  }
  return ans;
}
long long arr[1000005], pichle = 0, flip[1000005], drastic[1000005];
set<pair<long long, long long> > s;
int main() {
  long long n;
  cin >> n;
  long long now = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    if (i <= arr[i]) {
      pichle--;
      flip[arr[i] - i]++;
      drastic[n - i] = -(n - arr[i]) + (arr[i] - 1) + 1;
      flip[(n - i)]--;
    } else {
      pichle++;
      flip[n - i + arr[i]]++;
      drastic[n - i] = -(n - arr[i]) + (arr[i] - 1) + 1;
      flip[(n - i)]--;
    }
    now += abs(i - arr[i]);
  }
  s.insert(make_pair(now, 0));
  pichle += 2 * flip[0];
  for (long long k = 1; k < n; k++) {
    now += pichle + drastic[k - 1];
    pichle += 2 * flip[k];
    s.insert(make_pair(now, k));
  }
  cout << s.begin()->first << " " << s.begin()->second << '\n';
}

### Prompt

Please create a solution in Cpp 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;
long long gcd(long long n1, long long n2) {
  if (n2 == 0) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  if (exponent < 0) exponent += 1000000007LL - 1;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007LL;
    base = (base * base) % 1000000007LL;
    exponent /= 2;
  }
  return ans;
}
long long arr[1000005], pichle = 0, flip[1000005], drastic[1000005];
set<pair<long long, long long> > s;
int main() {
  long long n;
  cin >> n;
  long long now = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    if (i <= arr[i]) {
      pichle--;
      flip[arr[i] - i]++;
      drastic[n - i] = -(n - arr[i]) + (arr[i] - 1) + 1;
      flip[(n - i)]--;
    } else {
      pichle++;
      flip[n - i + arr[i]]++;
      drastic[n - i] = -(n - arr[i]) + (arr[i] - 1) + 1;
      flip[(n - i)]--;
    }
    now += abs(i - arr[i]);
  }
  s.insert(make_pair(now, 0));
  pichle += 2 * flip[0];
  for (long long k = 1; k < n; k++) {
    now += pichle + drastic[k - 1];
    pichle += 2 * flip[k];
    s.insert(make_pair(now, k));
  }
  cout << s.begin()->first << " " << s.begin()->second << '\n';
}

```