#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, a[N];
int dl[N], dr[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  long long ret = 0, ans = 0;
  int L = 0, R = 0;
  for (int i = 1; i <= n; ++i) {
    ret += abs(a[i] - i);
    if (a[i] > i) {
      ++L;
      --dl[a[i] - i];
      ++dr[a[i] - i];
    } else if (a[i] <= i) {
      ++R;
      if (a[i] != 1) {
        --dl[n - i + a[i]];
        ++dr[n - i + a[i]];
      }
    }
  }
  ans = ret;
  int pos = 0;
  for (int i = 1; i <= n - 1; ++i) {
    ret += R - L;
    R += dr[i];
    L += dl[i];
    if (a[n - i + 1] != 1) --R, ++L;
    ret -= abs(a[n - i + 1] - n - 1);
    ret += abs(a[n - i + 1] - 1);
    if (ret < ans) pos = i, ans = ret;
  }
  printf("%lld %d\n", ans, 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;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, a[N];
int dl[N], dr[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  long long ret = 0, ans = 0;
  int L = 0, R = 0;
  for (int i = 1; i <= n; ++i) {
    ret += abs(a[i] - i);
    if (a[i] > i) {
      ++L;
      --dl[a[i] - i];
      ++dr[a[i] - i];
    } else if (a[i] <= i) {
      ++R;
      if (a[i] != 1) {
        --dl[n - i + a[i]];
        ++dr[n - i + a[i]];
      }
    }
  }
  ans = ret;
  int pos = 0;
  for (int i = 1; i <= n - 1; ++i) {
    ret += R - L;
    R += dr[i];
    L += dl[i];
    if (a[n - i + 1] != 1) --R, ++L;
    ret -= abs(a[n - i + 1] - n - 1);
    ret += abs(a[n - i + 1] - 1);
    if (ret < ans) pos = i, ans = ret;
  }
  printf("%lld %d\n", ans, pos);
}

```