#include <bits/stdc++.h>
const int offset = 11E5;
int N;
int leftcnt[2400000], rightcnt[2400000];
int A[1100000];
int best = 0;
long long bans = 0;
long long ans = 0;
int main() {
  scanf("%d", &N);
  long long cur = 0;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    ans += abs(i - A[i]);
    cur += (A[i] <= i) ? 1 : -1;
    ++leftcnt[A[i] - i + offset];
  }
  best = 0;
  bans = ans;
  for (int i = 1; i <= N - 1; ++i) {
    ans += cur;
    ans = ans - (N + 1 - A[N + 1 - i]) + A[N + 1 - i] - 1;
    --leftcnt[A[N + 1 - i] - (N + 1 - i) + offset];
    ++rightcnt[A[N + 1 - i] - (N + 1 - i) + offset];
    cur -= 2;
    cur += leftcnt[i + offset] * 2;
    cur += rightcnt[i - N + offset] * 2;
    if (bans > ans) {
      best = i;
      bans = ans;
    }
  }
  printf("%lld %d\n", bans, best);
}

### Prompt

Please formulate 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>
const int offset = 11E5;
int N;
int leftcnt[2400000], rightcnt[2400000];
int A[1100000];
int best = 0;
long long bans = 0;
long long ans = 0;
int main() {
  scanf("%d", &N);
  long long cur = 0;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    ans += abs(i - A[i]);
    cur += (A[i] <= i) ? 1 : -1;
    ++leftcnt[A[i] - i + offset];
  }
  best = 0;
  bans = ans;
  for (int i = 1; i <= N - 1; ++i) {
    ans += cur;
    ans = ans - (N + 1 - A[N + 1 - i]) + A[N + 1 - i] - 1;
    --leftcnt[A[N + 1 - i] - (N + 1 - i) + offset];
    ++rightcnt[A[N + 1 - i] - (N + 1 - i) + offset];
    cur -= 2;
    cur += leftcnt[i + offset] * 2;
    cur += rightcnt[i - N + offset] * 2;
    if (bans > ans) {
      best = i;
      bans = ans;
    }
  }
  printf("%lld %d\n", bans, best);
}

```