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 long long inf = 1e18;
const long long maxn = 2e6 + 10;
long long num[maxn], n, neg, pos, tot;
inline long long hf(long long num) { return num + n - 1; }
inline long long hhf(long long num) { return num - n + 1; }
deque<long long> d, p;
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    p.push_back(x);
    num[hf(x - i - 1)]++;
  }
  for (long long i = 0; i < 2 * n - 1; i++) {
    d.push_back(num[i]);
    if (hhf(i) <= 0)
      neg += num[i];
    else
      pos += num[i];
    tot += num[i] * abs(hhf(i));
  }
  long long ans = tot, ansind = 0, New = tot;
  for (long long i = 1; i <= n - 1; i++) {
    neg -= (p[n - 1] - n <= 0);
    pos -= (p[n - 1] - n > 0);
    New += (neg - pos);
    New -= abs(p[n - 1] - n);
    New += abs(p[n - 1] - 1);
    d[hf(p[n - 1] - n)]--;
    pos -= d[hf(1)];
    neg += d[hf(1)];
    d.pop_front();
    d.push_back(0);
    d[hf(p[n - 1] - 1)]++;
    pos += (p[n - 1] - 1 > 0);
    neg += (p[n - 1] - 1 <= 0);
    p.push_front(p.back());
    p.pop_back();
    if (New < ans) {
      ans = New;
      ansind = i;
    }
  }
  cout << ans << " " << ansind;
}
