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;
long long n, ans, len, sumi, kl, k[2000100], p[2000100], sum, wapi[2000100],
    wai[2000100];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
    wapi[max(0ll, i - p[i])] += p[i];
    wai[max(0ll, i - p[i])] += i;
    ++k[max(0ll, i - p[i])];
    sum += p[i];
  }
  for (int i = 1; i <= n; ++i) {
    p[i + n] = p[i];
    wapi[max(0ll, i + n - p[i])] += p[i];
    wai[max(0ll, i + n - p[i])] += i + n;
    ++k[max(0ll, i + n - p[i])];
  }
  len = 1e18;
  long long sumwapi = 0, sumwai = 0;
  for (int i = n * 2; i > n + 1; --i) {
    sumwapi += wapi[i];
    sumwai += wai[i];
    kl += k[i];
  }
  for (int i = n + 1; i; --i) {
    sumwapi += wapi[i] - wapi[i + n];
    sumwai += wai[i] - wai[i + n];
    kl += k[i] + k[i + n];
    long long s = 2 * (sumwai - i * kl + kl) - 2 * sumwapi;
    wapi[max(0ll, i + n - 1 - p[i + n - 1])] -= p[i + n - 1];
    wai[max(0ll, i + n - 1 - p[i + n - 1])] -= i + n - 1;
    --k[max(0ll, i + n - 1 - p[i + n - 1])];
    if (max(0ll, i + n - 1 - p[i + n - 1]) >= i) {
      sumwapi -= p[i + n - 1];
      sumwai -= i + n - 1;
      --kl;
    }
    if (s < len) {
      len = s;
      ans = n + 1 - i;
    }
  }
  cout << len << ' ' << ans;
}
