#include <bits/stdc++.h>
using namespace std;
long long add[1000006], mul[1000006], ans[1000006];
int p[1000006], n;
void addRange_add(int x, int y, long long val) {
  if (x > y) return;
  add[x] += val;
  add[y + 1] -= val;
}
void addRange_mul(int x, int y, long long val) {
  if (x > y) return;
  mul[x] += val;
  mul[y + 1] -= val;
}
void process(int i) {
  if (i > p[i]) {
    addRange_add(n - i + 1, n - i + p[i], p[i]);
    addRange_add(0, n - i, -p[i]);
    addRange_add(n - i + p[i] + 1, n - 1, -p[i]);
    addRange_mul(n - i + 1, n - i + p[i], -1);
    addRange_add(n - i + 1, n - i + p[i], n - i);
    addRange_mul(0, n - i, 1);
    addRange_add(0, n - i, i);
    addRange_mul(n - i + p[i] + 1, n - 1, 1);
    addRange_add(n - i + p[i] + 1, n - 1, i - n);
  } else {
    addRange_add(0, p[i] - i, p[i]);
    addRange_add(n - i + 1, n - 1, p[i]);
    addRange_add(p[i] - i + 1, n - i, -p[i]);
    addRange_mul(0, p[i] - i, -1);
    addRange_add(0, p[i] - i, -i);
    addRange_mul(n - i + 1, n - 1, -1);
    addRange_add(n - i + 1, n - 1, n - i);
    addRange_mul(p[i] - i + 1, n - i, 1);
    addRange_add(p[i] - i + 1, n - i, i);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) process(i);
  for (int i = 0; i < n; i++) {
    ans[i] = add[i] + mul[i] * i;
    add[i + 1] += add[i];
    mul[i + 1] += mul[i];
  }
  int result = 0;
  for (int i = 0; i < n; i++)
    if (ans[i] < ans[result]) result = i;
  cout << ans[result] << ' ' << result;
}

### 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 add[1000006], mul[1000006], ans[1000006];
int p[1000006], n;
void addRange_add(int x, int y, long long val) {
  if (x > y) return;
  add[x] += val;
  add[y + 1] -= val;
}
void addRange_mul(int x, int y, long long val) {
  if (x > y) return;
  mul[x] += val;
  mul[y + 1] -= val;
}
void process(int i) {
  if (i > p[i]) {
    addRange_add(n - i + 1, n - i + p[i], p[i]);
    addRange_add(0, n - i, -p[i]);
    addRange_add(n - i + p[i] + 1, n - 1, -p[i]);
    addRange_mul(n - i + 1, n - i + p[i], -1);
    addRange_add(n - i + 1, n - i + p[i], n - i);
    addRange_mul(0, n - i, 1);
    addRange_add(0, n - i, i);
    addRange_mul(n - i + p[i] + 1, n - 1, 1);
    addRange_add(n - i + p[i] + 1, n - 1, i - n);
  } else {
    addRange_add(0, p[i] - i, p[i]);
    addRange_add(n - i + 1, n - 1, p[i]);
    addRange_add(p[i] - i + 1, n - i, -p[i]);
    addRange_mul(0, p[i] - i, -1);
    addRange_add(0, p[i] - i, -i);
    addRange_mul(n - i + 1, n - 1, -1);
    addRange_add(n - i + 1, n - 1, n - i);
    addRange_mul(p[i] - i + 1, n - i, 1);
    addRange_add(p[i] - i + 1, n - i, i);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) process(i);
  for (int i = 0; i < n; i++) {
    ans[i] = add[i] + mul[i] * i;
    add[i + 1] += add[i];
    mul[i + 1] += mul[i];
  }
  int result = 0;
  for (int i = 0; i < n; i++)
    if (ans[i] < ans[result]) result = i;
  cout << ans[result] << ' ' << result;
}

```