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;
using ll = long long;
using vi = vector<int>;
using db = double;
ll a[1000006];
ll p[1000006];
ll s[1000006];
void add(ll l, ll r, ll L, ll R) {
  if (L <= R) {
    a[l] += 1;
    a[r] -= 1;
    p[r] -= r - l;
    p[l] += L - 1;
    p[r] -= L - 1;
  } else {
    a[l] -= 1;
    a[r] += 1;
    p[r] += r - l;
    p[l] += L + 1;
    p[r] -= L + 1;
  }
}
int n;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    ll a;
    scanf("%lld", &a);
    --a;
    if (i >= a) {
      add(0, n - i, i - a, n - a);
      add(n - i, n - i + a, a - 0, 0);
      add(n - i + a, n, 0, i - a);
    } else {
      add(0, a - i, a - i, 0);
      add(a - i, n - i, 0, n - a);
      add(n - i, n, a, a - i);
    }
  }
  partial_sum(a, a + n, a);
  partial_sum(a, a + n, a);
  partial_sum(p, p + n, p);
  transform(a, a + n, p, s, plus<ll>());
  auto t = min_element(s, s + n);
  cout << *t << ' ' << t - s << endl;
  return 0;
}
