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 T>
void cmax(T& a, T b) {
  a = max(a, b);
}
template <typename T>
void cmin(T& a, T b) {
  a = min(a, b);
}
void _BG(const char* s) {}
template <typename T, typename... TT>
void _BG(const char* s, T a, TT... b) {
  for (int c = 0; *s && (c || *s != ','); ++s) {
    cerr << *s;
    switch (*s) {
      case '(':
      case '[':
      case '{':
        ++c;
        break;
      case ')':
      case ']':
      case '}':
        --c;
        break;
    }
  }
  cerr << " = " << a;
  if (*s) {
    cerr << ", ";
    _BG(++s, b...);
  } else
    cerr << endl;
}
bool RD(void) { return true; }
bool RD(char& a) { return scanf(" %c", &a) == 1; }
bool RD(char* a) { return scanf("%s", a) == 1; }
bool RD(double& a) { return scanf("%lf", &a) == 1; }
bool RD(int& a) { return scanf("%d", &a) == 1; }
bool RD(long long& a) { return scanf("%lld", &a) == 1; }
template <typename T, typename... TT>
bool RD(T& a, TT&... b) {
  return RD(a) && RD(b...);
}
void PT(const char a) { putchar(a); }
void PT(const char* a) { fputs(a, stdout); }
void PT(char* a) { fputs(a, stdout); }
void PT(const double a) { printf("%.16f", a); }
void PT(const int a) { printf("%d", a); }
void PT(const long long a) { printf("%lld", a); }
template <char sep = ' ', char end = '\n'>
void PL(void) {
  if (end) PT(end);
}
template <char sep = ' ', char end = '\n', typename T, typename... TT>
void PL(const T a, const TT... b) {
  PT(a);
  if (sizeof...(b) && sep) PT(sep);
  PL<sep, end>(b...);
}
const int N = 1e6 + 87;
int p[N], c[N], n, m;
int main() {
  RD(n);
  long long ans = 0, cur = 0;
  for (int i(1), _B((n) + 1); i < _B; ++i) {
    RD(p[i]);
    int v = p[i] - i;
    cur += abs(v);
    if (v >= 0) {
      ++m;
      c[v]++;
    }
  }
  ans = cur;
  int ak = 0;
  for (int k(1), _B(n); k < _B; ++k) {
    m -= c[k - 1];
    int j = n - k + 1;
    int v = p[j] - j;
    if (v >= k) {
      --m;
      --c[v];
    }
    cur += n - 1 - 2 * m + abs(p[j] - 1) - abs(p[j] - n);
    v = p[j] - 1 + k;
    if (v >= k) {
      ++m;
      if (v <= n - 2) ++c[v];
    }
    if (cur < ans) {
      ans = cur;
      ak = k;
    }
  }
  PL(ans, ak);
}
