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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
int N, A[(1000006)], g, l, G[(1000006)], L[(1000006)], P[(1000006)], a, W;
long long S, X;
int main(void) {
  scanf("%d", &N), ga(N, A + 1);
  for (int k(1); k < N + 1; k++)
    if (A[k] > k)
      ++g, S += a = A[k] - k, --G[a], ++L[a], P[N - k] += 2 * A[k] - 1 - N,
                ++G[N - k], --L[N - k];
    else
      ++l, S += k - A[k], a = N - k + A[k], --G[a], ++L[a],
                          P[N - k] += 2 * A[k] - 1 - N, ++G[N - k], --L[N - k];
  X = S;
  for (int k(0); k < N; k++) {
    l += L[k], g += G[k], S += P[k] + l - g + 1;
    if (S < X) X = S, W = k + 1;
  }
  printf("%lld %d\n", X, W);
  return 0;
}
