#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int sum = 0, t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return sum * t;
}
int n, p[1000005], pos;
long long step[1000005], now, S, ans;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1; i <= n; i++) {
    S += abs(p[i] - i);
    if (p[i] >= i) {
      step[0] += -1;
      step[p[i] - i] += 1 + 1;
      step[n - i] += abs(p[i] - 1) - abs(p[i] - n) - 1;
      step[n - i + 1] += -1 - abs(p[i] - 1) + abs(p[i] - n);
    } else if (p[i] < i) {
      step[0] += 1;
      step[n - i] += abs(p[i] - 1) - abs(p[i] - n) - 1;
      step[n - i + 1] += -1 - abs(p[i] - 1) + abs(p[i] - n);
      step[p[i] + n - i] += 1 + 1;
    }
  }
  ans = S;
  pos = 0;
  for (int k = 0; k < n; k++) {
    S += now;
    if (S < ans) ans = S, pos = k;
    now += step[k];
  }
  printf("%lld %d\n", ans, pos);
}

### Prompt

Create a solution in cpp for 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;
inline int read() {
  int sum = 0, t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return sum * t;
}
int n, p[1000005], pos;
long long step[1000005], now, S, ans;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1; i <= n; i++) {
    S += abs(p[i] - i);
    if (p[i] >= i) {
      step[0] += -1;
      step[p[i] - i] += 1 + 1;
      step[n - i] += abs(p[i] - 1) - abs(p[i] - n) - 1;
      step[n - i + 1] += -1 - abs(p[i] - 1) + abs(p[i] - n);
    } else if (p[i] < i) {
      step[0] += 1;
      step[n - i] += abs(p[i] - 1) - abs(p[i] - n) - 1;
      step[n - i + 1] += -1 - abs(p[i] - 1) + abs(p[i] - n);
      step[p[i] + n - i] += 1 + 1;
    }
  }
  ans = S;
  pos = 0;
  for (int k = 0; k < n; k++) {
    S += now;
    if (S < ans) ans = S, pos = k;
    now += step[k];
  }
  printf("%lld %d\n", ans, pos);
}

```