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;
const int MAXN = 1000000 + 10;
int N;
int P[MAXN], PMI[MAXN];
int H[MAXN * 4];
long long A[MAXN];
void Work() {
  scanf("%d", &N);
  long long Init = 0;
  long long Neg = 0, NonNeg = 0;
  long long Ans = 100000000000000LL, AnsPos = -1;
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
    P[i]--;
    PMI[i] = P[i] - i;
    H[PMI[i] + MAXN * 2]++;
    Init += (long long)abs(PMI[i]);
    if (PMI[i] < 0)
      Neg++;
    else
      NonNeg++;
  }
  Ans = Init;
  AnsPos = 0;
  A[0] = Ans;
  for (int k = 0; k < N - 1; k++) {
    long long OldPMIk = PMI[k];
    long long NewPMIk = PMI[k] - N;
    H[OldPMIk + MAXN * 2]--;
    H[NewPMIk + MAXN * 2]++;
    Init -= abs(P[k]);
    Init += abs(P[k] - N);
    if (P[k] < 0)
      Neg--;
    else
      NonNeg--;
    if (P[k] - N < 0)
      Neg++;
    else
      NonNeg++;
    Init -= Neg;
    Init += NonNeg;
    Neg -= H[-k - 1 + MAXN * 2];
    NonNeg += H[-k - 1 + MAXN * 2];
    if (Init < Ans) {
      Ans = Init;
      AnsPos = N - (k + 1);
    }
    A[N - (k + 1)] = Init;
  }
  printf("%I64d %I64d\n", Ans, AnsPos);
}
int main() {
  Work();
  return 0;
}
