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 INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
int v[1000100];
int v2[1000100];
int n;
long long int S = 0;
int antes = 0, depois = 0;
int mapa[1000100];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    v[i]--;
    S += abs(v[i] - i);
    if (i < n - 1) {
      antes += ((v[i] - i) > 0);
      depois += ((v[i] - i) <= 0);
    }
  }
  for (int i = 0; i < n; i++) {
    int s = v[i] - i;
    if (s < 0) s += n;
    v2[i] = s;
    mapa[v2[i]]++;
  }
  int j = n - 1;
  long long int r = S;
  int idx = 0;
  for (int i = 1; i < n; i++) {
    S = S - abs(v[j] - n + 1);
    S = S + v[j] - antes + depois;
    antes = antes - mapa[i] + 1 - (v[j] == 0) - (v[j - 1] - n + 1 > 0);
    depois = depois + mapa[i] - 1 + (v[j] != 0) - (v[j - 1] - n + 1 <= 0);
    j--;
    if (S < r) {
      r = S;
      idx = i;
    }
  }
  cout << r << " " << idx << endl;
  return 0;
}
