#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, p[2000100], q[2000100], l, r, s[2000100], ct1, ct2, Ans, nw, ps;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) q[i + n] = p[i] = read();
  Ans = 10000000000000ll;
  l = n + 1;
  r = n + n;
  for (long long i = 1; i <= n; i++) {
    nw += abs(p[i] - i);
    if (p[i] > i)
      s[p[i] - i]++, ct1++;
    else
      ct2++;
  }
  Ans = min(Ans, nw);
  for (long long i = 1; i <= n; i++) {
    nw += ct2 - ct1;
    nw -= abs(q[r] - n - 1);
    nw += abs(q[r] - 1);
    if (Ans > nw) Ans = nw, ps = i;
    ct2--;
    if (q[r] > 1)
      s[i + q[r] - 1]++, ct1++;
    else
      ct2++;
    ct1 -= s[i];
    ct2 += s[i];
    q[--l] = q[r];
    r--;
  }
  cout << Ans << " " << ps << endl;
  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 ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, p[2000100], q[2000100], l, r, s[2000100], ct1, ct2, Ans, nw, ps;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) q[i + n] = p[i] = read();
  Ans = 10000000000000ll;
  l = n + 1;
  r = n + n;
  for (long long i = 1; i <= n; i++) {
    nw += abs(p[i] - i);
    if (p[i] > i)
      s[p[i] - i]++, ct1++;
    else
      ct2++;
  }
  Ans = min(Ans, nw);
  for (long long i = 1; i <= n; i++) {
    nw += ct2 - ct1;
    nw -= abs(q[r] - n - 1);
    nw += abs(q[r] - 1);
    if (Ans > nw) Ans = nw, ps = i;
    ct2--;
    if (q[r] > 1)
      s[i + q[r] - 1]++, ct1++;
    else
      ct2++;
    ct1 -= s[i];
    ct2 += s[i];
    q[--l] = q[r];
    r--;
  }
  cout << Ans << " " << ps << endl;
  return 0;
}

```