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;
vector<long long> pri(1000009, 0);
vector<long long> sec(1000009, 0);
void add_val(long long l, long long r, long long val) {
  pri[l] += val;
  pri[r + 1] -= val;
}
void add_prog(long long l, long long r, long long val) {
  pri[r + 1] -= val * (r - l + 1);
  sec[l] += val;
  sec[r + 1] -= val;
}
void add_comb(long long l, long long r, long long val1, long long val2) {
  add_val(l, r, val1);
  add_prog(l, r, val2);
}
void final_proc() {
  for (long long i = 1; i < (1000009); i++) sec[i] += sec[i - 1];
  for (long long i = 1; i < (1000009); i++) pri[i] += sec[i - 1];
  for (long long i = 1; i < (1000009); i++) pri[i] += pri[i - 1];
}
int main() {
  long long(n);
  scanf("%lld", &(n));
  ;
  vector<long long> A((n));
  for (long long i = 0; i < n; i++) scanf("%lld", &A[i]);
  ;
  for (long long i = 0; i < (n); i++) {
    A[i]--;
    if (A[i] < i) {
      add_comb(0, n - 1 - i, i - A[i], 1);
      add_comb(n - i, n - i + A[i], A[i], -1);
      add_comb(n - i + A[i] + 1, n - 1, 1, 1);
    } else if (A[i] >= i) {
      add_comb(0, A[i] - i, A[i] - i, -1);
      add_comb(A[i] - i + 1, n - 1 - i, 1, 1);
      add_comb(n - i, n - 1, A[i], -1);
    }
  }
  final_proc();
  long long idx = -1, mini = (long long)1e18;
  for (long long i = 0; i < (n); i++) {
    if (mini > pri[i]) {
      mini = pri[i];
      idx = i;
    }
  }
  printf("%lld %lld\n", mini, idx);
  return 0;
}
