#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1LL << 50;
const double eps = 1e-13;
long long sums[1000009];
long long l[1000009], r[1000009];
int d[1000009];
vector<int> p[4000009], p2[4000009];
int main() {
  int(n);
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &(d[i]));
    p2[d[i] - 1 + (n - 1 - i) + 2000000].push_back(i);
    p[d[i] - n - i + 2000000].push_back(i);
  }
  int pv = 0, nv = 0;
  for (int i = 0; i < (n); ++i) {
    int c = d[i] - n;
    l[i + 1] = l[i] + abs(c) + pv - nv;
    if (c >= 0)
      pv++;
    else
      nv++;
    for (int j = 0; j < (((int)(p[-i + 2000000]).size())); ++j) {
      if (p[-i + 2000000][j] >= i) break;
      pv++;
      nv--;
    }
  }
  pv = 0, nv = 0;
  for (int i = n - 1; i >= 0; i--) {
    int c = d[i] - 1;
    r[i] = r[i + 1] + abs(c) + nv - pv;
    if (c <= 0)
      nv++;
    else
      pv++;
    for (int j = ((int)(p2[n - 1 - i + 2000000]).size()) - 1; j >= 0; j--) {
      if (p2[n - 1 - i + 2000000][j] <= i) break;
      nv++;
      pv--;
    }
  }
  long long best = 1LL << 60;
  int bi = 0;
  for (int i = (1); i < (n + 1); ++i) {
    long long cur = l[i] + r[i];
    if (cur < best) {
      best = cur;
      bi = n - i;
    }
  }
  printf("%I64d %d", best, bi % n);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long MOD = 1e9 + 7;
const long long INF = 1LL << 50;
const double eps = 1e-13;
long long sums[1000009];
long long l[1000009], r[1000009];
int d[1000009];
vector<int> p[4000009], p2[4000009];
int main() {
  int(n);
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &(d[i]));
    p2[d[i] - 1 + (n - 1 - i) + 2000000].push_back(i);
    p[d[i] - n - i + 2000000].push_back(i);
  }
  int pv = 0, nv = 0;
  for (int i = 0; i < (n); ++i) {
    int c = d[i] - n;
    l[i + 1] = l[i] + abs(c) + pv - nv;
    if (c >= 0)
      pv++;
    else
      nv++;
    for (int j = 0; j < (((int)(p[-i + 2000000]).size())); ++j) {
      if (p[-i + 2000000][j] >= i) break;
      pv++;
      nv--;
    }
  }
  pv = 0, nv = 0;
  for (int i = n - 1; i >= 0; i--) {
    int c = d[i] - 1;
    r[i] = r[i + 1] + abs(c) + nv - pv;
    if (c <= 0)
      nv++;
    else
      pv++;
    for (int j = ((int)(p2[n - 1 - i + 2000000]).size()) - 1; j >= 0; j--) {
      if (p2[n - 1 - i + 2000000][j] <= i) break;
      nv++;
      pv--;
    }
  }
  long long best = 1LL << 60;
  int bi = 0;
  for (int i = (1); i < (n + 1); ++i) {
    long long cur = l[i] + r[i];
    if (cur < best) {
      best = cur;
      bi = n - i;
    }
  }
  printf("%I64d %d", best, bi % n);
}

```