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;
int const N = 1000100;
int a[N], s[N * 2];
void bru(int n) {
  int ret = 0x3f3f3f3f;
  for (int i = 1; i <= (n); ++i) {
    int now = 0;
    for (int j = 0; j < (n); ++j) {
      int t = i + j;
      if (t > n) t -= n;
      now += max(j + 1 - a[t], a[t] - j - 1);
    }
    ret = min(ret, now);
  }
  cout << ret << endl;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= (n); ++i) scanf("%d", &a[i]);
  long long up = 0, down = 0;
  long long now = 0, ret = 0, rc = 0;
  for (int i = 1; i <= (n); ++i) {
    if (a[i] >= i)
      ++up;
    else {
      ++down;
      s[i - a[i]]++;
    }
    now += max(a[i] - i, i - a[i]);
  }
  ret = now;
  int step = 0;
  for (int i = 1; i <= n - 1; ++i) {
    ++step;
    --up;
    now -= a[i] - 1;
    now += n - a[i];
    now += up;
    now -= down;
    s[step + n - a[i]]++;
    up += s[step];
    down -= s[step];
    ++down;
    if (now < ret) {
      ret = now;
      rc = n - i;
    }
  }
  cout << ret << " " << rc << endl;
  return 0;
}
