#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char a = getchar();
  while (a < '0' || a > '9') {
    if (a == '-') f = -1;
    a = getchar();
  }
  while (a >= '0' && a <= '9') x = x * 10 + a - '0', a = getchar();
  return x * f;
}
int n, p[1000000 + 5], neg, delta, num;
long long ans, now;
priority_queue<int> q;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) p[i] = read(), ans += abs(p[i] - i);
  now = ans;
  num = n;
  for (int i = 1; i <= n; i++)
    if (p[i] - i <= 0)
      neg++;
    else
      q.push(i - p[i]);
  for (int i = n - 1; i; i--) {
    int tmp = neg - q.size();
    if (p[i + 1] - n <= 0)
      tmp--;
    else
      tmp++;
    now = now - abs(p[i + 1] - n) + abs(p[i + 1] - 1) + tmp;
    if (ans > now) ans = now, num = i;
    delta++;
    while (!q.empty()) {
      int x = -q.top();
      if (x - delta > 0) break;
      neg++;
      q.pop();
    }
    if (p[i + 1] - n <= 0 && p[i + 1] - 1 > 0)
      neg--, q.push(-p[i + 1] + 1 - delta);
  }
  printf("%I64d %d\n", ans, n - num);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char a = getchar();
  while (a < '0' || a > '9') {
    if (a == '-') f = -1;
    a = getchar();
  }
  while (a >= '0' && a <= '9') x = x * 10 + a - '0', a = getchar();
  return x * f;
}
int n, p[1000000 + 5], neg, delta, num;
long long ans, now;
priority_queue<int> q;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) p[i] = read(), ans += abs(p[i] - i);
  now = ans;
  num = n;
  for (int i = 1; i <= n; i++)
    if (p[i] - i <= 0)
      neg++;
    else
      q.push(i - p[i]);
  for (int i = n - 1; i; i--) {
    int tmp = neg - q.size();
    if (p[i + 1] - n <= 0)
      tmp--;
    else
      tmp++;
    now = now - abs(p[i + 1] - n) + abs(p[i + 1] - 1) + tmp;
    if (ans > now) ans = now, num = i;
    delta++;
    while (!q.empty()) {
      int x = -q.top();
      if (x - delta > 0) break;
      neg++;
      q.pop();
    }
    if (p[i + 1] - n <= 0 && p[i + 1] - 1 > 0)
      neg--, q.push(-p[i + 1] + 1 - delta);
  }
  printf("%I64d %d\n", ans, n - num);
  return 0;
}

```