#include <bits/stdc++.h>
using namespace std;
int IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 2000000 + 19;
int P[N], pool[4 * N], *cnt = pool + 2 * N;
int n, cl, cr, plu, pos;
long long res = 1ll << 60, tmp;
int main() {
  n = IN();
  for (int i = 1; i < n + 1; i++) P[i] = IN();
  for (int i = 1; i < n + 1; i++) {
    cnt[P[i] - i]++;
    if (P[i] - i < 0)
      cl++;
    else
      cr++;
    tmp += abs(P[i] - i);
  }
  for (int i = n; i; i--) {
    if (tmp < res) res = tmp, pos = n - i;
    cnt[P[i] - n - plu]--;
    tmp -= abs(P[i] - n);
    if (P[i] - n < 0)
      cl--;
    else
      cr--;
    cl += cnt[-plu];
    cr -= cnt[-plu];
    plu--;
    tmp += cl;
    tmp -= cr;
    cnt[P[i] - 1 - plu]++;
    tmp += abs(P[i] - 1);
    if (P[i] - 1 < 0)
      cl++;
    else
      cr++;
  }
  printf("%I64d %d\n", res, pos);
}

### Prompt

Generate 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;
int IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 2000000 + 19;
int P[N], pool[4 * N], *cnt = pool + 2 * N;
int n, cl, cr, plu, pos;
long long res = 1ll << 60, tmp;
int main() {
  n = IN();
  for (int i = 1; i < n + 1; i++) P[i] = IN();
  for (int i = 1; i < n + 1; i++) {
    cnt[P[i] - i]++;
    if (P[i] - i < 0)
      cl++;
    else
      cr++;
    tmp += abs(P[i] - i);
  }
  for (int i = n; i; i--) {
    if (tmp < res) res = tmp, pos = n - i;
    cnt[P[i] - n - plu]--;
    tmp -= abs(P[i] - n);
    if (P[i] - n < 0)
      cl--;
    else
      cr--;
    cl += cnt[-plu];
    cr -= cnt[-plu];
    plu--;
    tmp += cl;
    tmp -= cr;
    cnt[P[i] - 1 - plu]++;
    tmp += abs(P[i] - 1);
    if (P[i] - 1 < 0)
      cl++;
    else
      cr++;
  }
  printf("%I64d %d\n", res, pos);
}

```