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 inf = 0x3f3f3f3f;
const double eps = 1e-8;
int p[1000005];
int d[1000005];
int cnt[1000005];
int main() {
  ios_base::sync_with_stdio(0);
  int n, c, pos, neg;
  long long int mx, cur;
  while (cin >> n) {
    memset(cnt, 0, sizeof(cnt));
    mx = c = pos = neg = 0;
    for (int i = 1; i <= n; ++i) {
      cin >> p[i];
      d[i] = i - p[i];
      if (d[i] >= 0)
        ++pos;
      else {
        ++neg;
        cnt[-d[i]]++;
      }
      mx += abs(d[i]);
    }
    cur = mx;
    for (int i = 1; i < n; ++i) {
      --pos;
      cur -= n - p[n + 1 - i];
      cur += p[n + 1 - i] - 1;
      cur += pos;
      cur -= neg;
      ++neg;
      cnt[p[n + 1 - i] + i - 1]++;
      neg -= cnt[i];
      pos += cnt[i];
      if (cur < mx) {
        mx = cur;
        c = i;
      }
    }
    cout << mx << " " << c << "\n";
  }
  return 0;
}
