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 a[N], n;
long long l[N], r[N];
pair<long long, int> tr[N];
void add(int x, pair<long long, int> v) {
  for (int i = x; i <= 2 * n; i += (i & -i))
    tr[i].first += v.first, tr[i].second += v.second;
}
pair<long long, int> sum(int x) {
  pair<long long, int> res = {0, 0};
  for (int i = x; i >= 1; i -= (i & -i))
    res.first += tr[i].first, res.second += tr[i].second;
  return res;
}
int main() {
  scanf("%d ", &n);
  for (int i = 1; i <= n; i++) scanf("%d ", a + i);
  for (int i = 1; i <= n; i++) a[i] = a[i] - i;
  for (int i = 1; i <= n; i++) {
    add(a[i] + n, {a[i], 1});
    long long t = sum(2 * n).first - sum(2 * n - i - 1).first,
              temp = sum(2 * n).second - sum(2 * n - i - 1).second;
    l[i] = t - (long long)(n - i) * temp;
  }
  for (int i = 1; i <= 2 * n; i++) tr[i] = {0, 0};
  for (int i = n, j = 1; i >= 1; i--, j++) {
    add(a[i] + n, {a[i], 1});
    long long t = sum(2 * n).first - sum(j - 1).first,
              temp = sum(2 * n).second - sum(j - 1).second;
    r[i] = t + temp * (n - j);
  }
  long long Min = 1e18, temp = -1;
  for (int i = n, k = 0; i >= 1; i--, k++) {
    if (Min > (l[i] + r[i + 1]) * 2) Min = (l[i] + r[i + 1]) * 2, temp = k;
  }
  printf("%lld %lld", Min, temp);
  return 0;
}
