#include <bits/stdc++.h>
using namespace std;
namespace ioput {
template <typename T>
T read() {
  T x = 0, f = 1;
  char c = getchar_unlocked();
  while ((c < '0') || (c > '9')) {
    if (c == '-') f = -1;
    c = getchar_unlocked();
  }
  while ((c >= '0') && (c <= '9'))
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar_unlocked();
  return x * f;
}
template <typename T>
void write(T x, char c) {
  static char t[20];
  static int tlen;
  t[tlen = 1] = c;
  if (x < 0) putchar_unlocked('-'), x = -x;
  do t[++tlen] = x % 10 + '0', x /= 10;
  while (x);
  while (tlen) putchar_unlocked(t[tlen--]);
}
}  // namespace ioput
using namespace ioput;
const int maxn = 10000005;
int p[maxn];
long long c[maxn], ans[maxn];
int main() {
  int n = read<int>();
  for (int i = 1, iend = n; i <= iend; ++i) {
    int t;
    p[i] = read<int>();
    ans[n - i + 1] = 2 * p[i] - n - 1;
    if (p[i] >= i) {
      t = p[i] - i;
      --c[1], ++c[t + 1];
      ++c[t + 1], --c[n - i + 1];
      --c[n - i + 2];
      ans[0] += p[i] - i;
    } else {
      t = p[i] - i + n;
      ++c[1], --c[n - i + 1];
      --c[n - i + 2], ++c[t + 1];
      ++c[t + 1];
      ans[0] += i - p[i];
    }
  }
  for (int i = 1, iend = n - 1; i <= iend; ++i)
    c[i] += c[i - 1], ans[i] += ans[i - 1] + c[i];
  int res = 0;
  for (int i = 1, iend = n - 1; i <= iend; ++i)
    if (ans[i] < ans[res]) res = i;
  write(ans[res], ' '), write(res, '\n');
  return 0;
}

### Prompt

Please formulate 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;
namespace ioput {
template <typename T>
T read() {
  T x = 0, f = 1;
  char c = getchar_unlocked();
  while ((c < '0') || (c > '9')) {
    if (c == '-') f = -1;
    c = getchar_unlocked();
  }
  while ((c >= '0') && (c <= '9'))
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar_unlocked();
  return x * f;
}
template <typename T>
void write(T x, char c) {
  static char t[20];
  static int tlen;
  t[tlen = 1] = c;
  if (x < 0) putchar_unlocked('-'), x = -x;
  do t[++tlen] = x % 10 + '0', x /= 10;
  while (x);
  while (tlen) putchar_unlocked(t[tlen--]);
}
}  // namespace ioput
using namespace ioput;
const int maxn = 10000005;
int p[maxn];
long long c[maxn], ans[maxn];
int main() {
  int n = read<int>();
  for (int i = 1, iend = n; i <= iend; ++i) {
    int t;
    p[i] = read<int>();
    ans[n - i + 1] = 2 * p[i] - n - 1;
    if (p[i] >= i) {
      t = p[i] - i;
      --c[1], ++c[t + 1];
      ++c[t + 1], --c[n - i + 1];
      --c[n - i + 2];
      ans[0] += p[i] - i;
    } else {
      t = p[i] - i + n;
      ++c[1], --c[n - i + 1];
      --c[n - i + 2], ++c[t + 1];
      ++c[t + 1];
      ans[0] += i - p[i];
    }
  }
  for (int i = 1, iend = n - 1; i <= iend; ++i)
    c[i] += c[i - 1], ans[i] += ans[i - 1] + c[i];
  int res = 0;
  for (int i = 1, iend = n - 1; i <= iend; ++i)
    if (ans[i] < ans[res]) res = i;
  write(ans[res], ' '), write(res, '\n');
  return 0;
}

```