#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
long long A[N], B[N];
int x[N];
int n;
void addin(int l, int r, int k, int b) {
  A[l] += b;
  B[l] += k;
  A[r + 1] -= b;
  B[r + 1] -= k;
}
void add(int l, int r, int where, int w) {
  if (l > r) return;
  int wherel = (l + where - 1) % n + 1, wherer = (r + where - 1) % n + 1;
  int wl = abs(wherel - w), wr = abs(wherer - w);
  if (wl <= wr)
    addin(l, r, wl - l, 1);
  else
    addin(l, r, wl + l, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) {
    int k1 = n - i, k2 = (x[i] - i + n) % n;
    if (k1 > k2) swap(k1, k2);
    add(0, k1, i, x[i]);
    add(k1 + 1, k2, i, x[i]);
    add(k2 + 1, n - 1, i, x[i]);
  }
  for (int i = 1; i <= n; i++) A[i] += A[i - 1], B[i] += B[i - 1];
  long long ans = 1e18;
  int where = 0;
  for (int i = 0; i < n; i++) {
    long long w = 1ll * A[i] * i + B[i];
    if (w < ans) ans = w, where = i;
  }
  cout << ans << " " << where << 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;
const int N = 1e6 + 10;
long long A[N], B[N];
int x[N];
int n;
void addin(int l, int r, int k, int b) {
  A[l] += b;
  B[l] += k;
  A[r + 1] -= b;
  B[r + 1] -= k;
}
void add(int l, int r, int where, int w) {
  if (l > r) return;
  int wherel = (l + where - 1) % n + 1, wherer = (r + where - 1) % n + 1;
  int wl = abs(wherel - w), wr = abs(wherer - w);
  if (wl <= wr)
    addin(l, r, wl - l, 1);
  else
    addin(l, r, wl + l, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) {
    int k1 = n - i, k2 = (x[i] - i + n) % n;
    if (k1 > k2) swap(k1, k2);
    add(0, k1, i, x[i]);
    add(k1 + 1, k2, i, x[i]);
    add(k2 + 1, n - 1, i, x[i]);
  }
  for (int i = 1; i <= n; i++) A[i] += A[i - 1], B[i] += B[i - 1];
  long long ans = 1e18;
  int where = 0;
  for (int i = 0; i < n; i++) {
    long long w = 1ll * A[i] * i + B[i];
    if (w < ans) ans = w, where = i;
  }
  cout << ans << " " << where << endl;
  return 0;
}

```