#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6;
int a[maxn + 50], b[maxn + 50];
int x[maxn * 2 + 50], y[maxn * 2 + 50];
int numx[3], numy[3];
int n;
long long ans, s;
int ansi;
int getnum(int x, int zero) {
  if (x == zero) return 0;
  if (x > zero) return 1;
  return 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) b[i] = a[i] - i, ++x[b[i] + n], ans += abs(b[i]);
  ansi = 0;
  s = ans;
  for (int i = 1; i <= n; ++i) ++numx[getnum(b[i], 0)];
  for (int i = n; i >= 2; --i) {
    --x[b[i] + n];
    --numx[getnum(b[i], n - i)];
    numx[2] += numx[0];
    numx[0] = x[n - i + 1 + n];
    numx[1] -= x[n - i + 1 + n];
    s += numx[2];
    s -= numx[0] + numx[1];
    s -= abs(a[i] - n);
    s += abs(1 - a[i]);
    numy[2] += numy[0];
    numy[0] = y[n - i + n];
    numy[1] -= y[n - i + n];
    s += numy[2];
    s -= numy[0] + numy[1];
    ++y[a[i] - 1 + n - i + n];
    ++numy[getnum(a[i] - 1 + n - i, n - i)];
    if (s < ans) {
      ans = s;
      ansi = n - i + 1;
    }
  }
  printf("%lld %d", ans, ansi);
  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 maxn = 1e6;
int a[maxn + 50], b[maxn + 50];
int x[maxn * 2 + 50], y[maxn * 2 + 50];
int numx[3], numy[3];
int n;
long long ans, s;
int ansi;
int getnum(int x, int zero) {
  if (x == zero) return 0;
  if (x > zero) return 1;
  return 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) b[i] = a[i] - i, ++x[b[i] + n], ans += abs(b[i]);
  ansi = 0;
  s = ans;
  for (int i = 1; i <= n; ++i) ++numx[getnum(b[i], 0)];
  for (int i = n; i >= 2; --i) {
    --x[b[i] + n];
    --numx[getnum(b[i], n - i)];
    numx[2] += numx[0];
    numx[0] = x[n - i + 1 + n];
    numx[1] -= x[n - i + 1 + n];
    s += numx[2];
    s -= numx[0] + numx[1];
    s -= abs(a[i] - n);
    s += abs(1 - a[i]);
    numy[2] += numy[0];
    numy[0] = y[n - i + n];
    numy[1] -= y[n - i + n];
    s += numy[2];
    s -= numy[0] + numy[1];
    ++y[a[i] - 1 + n - i + n];
    ++numy[getnum(a[i] - 1 + n - i, n - i)];
    if (s < ans) {
      ans = s;
      ansi = n - i + 1;
    }
  }
  printf("%lld %d", ans, ansi);
  return 0;
}

```