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;
template <typename T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
const int N = 1000001;
long long _a[3 * N], *a = _a + 2 * N;
long long pos[N];
void solve() {
  long long n, v;
  cin >> n;
  long long less = 0;
  long long dif = 0;
  for (int i = 1; i <= (n); ++i) {
    cin >> v;
    a[i - v]++;
    pos[i] = i - v;
    less += i < v;
    dif += abs(i - v);
  }
  long long mini = 0;
  long long ans = dif;
  long long cur = 0;
  for (int i = 1; i <= (n - 1); ++i) {
    dif += 2 * (cur - pos[n + 1 - i]) + n;
    a[pos[n + 1 - i] - n]++;
    less++;
    dif += n - 2 * less;
    if (Min(ans, dif)) mini = i;
    less -= a[--cur];
  }
  cout << ans << ' ' << mini << '\n';
}
void init() {}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  init();
  solve();
  return 0;
}
