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.
#include <bits/stdc++.h>
using namespace std;
template <typename G>
inline void read(G &x) {
  x = 0;
  char c;
  while (!isdigit(c = getchar()))
    ;
  do {
    x = x * 10 + c - '0';
  } while (isdigit(c = getchar()));
}
template <typename G>
inline void write(G x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(a, min(b, c, d));
}
template <class T>
inline T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(a, max(b, c, d));
}
const int MAXN = 1e6 + 7;
const int base = 1e9 + 7;
const int N = 2e6 + 7;
int n, m;
int p[MAXN], bit[2 * MAXN];
long long res;
void upd(int x, int value) {
  while (x <= 2 * m) {
    bit[x] += value;
    x += x & (-x);
  }
}
int get(int x) {
  int res = 0;
  while (x) {
    res += bit[x];
    x -= x & (-x);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read(n);
  for (int i = (1); i <= (n); ++i) read(p[i]);
  m = n;
  for (int i = (1); i <= (n); ++i) {
    upd(p[i] - i + m, 1);
    res += abs(p[i] - i);
  }
  long long mii = res;
  int ans = 0;
  for (int k = (1); k <= (n); ++k) {
    int i = n - k + 1;
    res -= abs(p[i] - n);
    res += abs(p[i] - 1);
    upd(p[i] - n + m, -1);
    int x = get(m + k - 1);
    res += x;
    res -= (n - 1 - x);
    upd(p[i] - 1 + m + k, 1);
    if (res < mii) {
      ans = k;
      mii = res;
    }
  }
  cout << mii << ' ' << ans << '\n';
  return 0;
}
