#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void sc(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
int p[1000001], sum[1000001];
int main() {
  int i, k, n, pnum = 0, nnum = 0;
  long long mini, tot = 0;
  sc(n);
  for (i = 1; i < n + 1; i++) {
    sc(p[i]);
    if (p[i] - i >= 0) {
      sum[p[i] - i]++;
      pnum++;
    } else
      nnum++;
    tot += abs(p[i] - i);
  }
  mini = tot, k = 0;
  for (i = 1; i < n; i++) {
    if (p[n - i + 1] != n)
      nnum--;
    else {
      pnum--;
      sum[p[n - i + 1] - n + i - 1]--;
    }
    if (p[n - i + 1] < n - i + 1) sum[p[n - i + 1] + i - 1]++;
    pnum -= sum[i - 1];
    nnum += sum[i - 1];
    sum[i - 1] = 0;
    tot -= abs(p[n - i + 1] - n);
    tot += p[n - i + 1] - 1;
    tot -= (pnum - nnum);
    pnum++;
    if (tot < mini) mini = tot, k = i;
  }
  printf("%lld %d\n", mini, k);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long mod = 1e9 + 7;
void sc(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
int p[1000001], sum[1000001];
int main() {
  int i, k, n, pnum = 0, nnum = 0;
  long long mini, tot = 0;
  sc(n);
  for (i = 1; i < n + 1; i++) {
    sc(p[i]);
    if (p[i] - i >= 0) {
      sum[p[i] - i]++;
      pnum++;
    } else
      nnum++;
    tot += abs(p[i] - i);
  }
  mini = tot, k = 0;
  for (i = 1; i < n; i++) {
    if (p[n - i + 1] != n)
      nnum--;
    else {
      pnum--;
      sum[p[n - i + 1] - n + i - 1]--;
    }
    if (p[n - i + 1] < n - i + 1) sum[p[n - i + 1] + i - 1]++;
    pnum -= sum[i - 1];
    nnum += sum[i - 1];
    sum[i - 1] = 0;
    tot -= abs(p[n - i + 1] - n);
    tot += p[n - i + 1] - 1;
    tot -= (pnum - nnum);
    pnum++;
    if (tot < mini) mini = tot, k = i;
  }
  printf("%lld %d\n", mini, k);
  return 0;
}

```