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 N = 2e6 + 10;
int z[N], f[N], p[N], n;
int sum[N * 4];
void update(int i, int v) {
  for (; i < n * 4; i += (i & -i)) sum[i] += v;
}
int getsum(int i) {
  int ans = 0;
  for (; i; i -= (i & -i)) ans += sum[i];
  return ans;
}
int main() {
  memset(sum, 0, sizeof(sum));
  scanf("%d", &n);
  long long int tmp = 0, tem = 0;
  int ans = 0, dd = n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    tmp += ((p[i] - i) > 0 ? (p[i] - i) : -(p[i] - i));
    update(p[i] - i + n, 1);
  }
  tem = tmp;
  for (int i = 2; i <= n; i++) {
    tem -= ((p[n - i + 2] - n) > 0 ? (p[n - i + 2] - n) : -(p[n - i + 2] - n));
    tem += ((p[n - i + 2] - 1) > 0 ? (p[n - i + 2] - 1) : -(p[n - i + 2] - 1));
    update(p[n - i + 2] - (n - i + 2) + n, -1);
    int t = getsum(dd);
    tem += t;
    tem -= n - 1 - t;
    dd++;
    update(p[n - i + 2] - 1 + dd, 1);
    if (tem < tmp) {
      tmp = tem;
      ans = i - 1;
    }
  }
  printf("%lld %d\n", tmp, ans);
  return 0;
}
