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 MAXN = 1000010;
int n, a[MAXN << 1];
void input() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    a[i + n] = a[i];
  }
}
vector<int> c[MAXN << 1];
void prelude() {
  for (int i = 1; i <= n + n; ++i) {
    int idx = n - i + a[i];
    if (idx < n && idx > 0) c[idx].push_back(i);
  }
}
int b[MAXN << 1];
void solve() {
  long long ans = 0, sum = 0, best = 0;
  int nega = 0, posi = 0;
  for (int i = n + 1; i <= n + n; ++i) {
    int num = a[i] - (i - n);
    if (num <= 0)
      b[i] = -1, ++nega;
    else
      b[i] = 1, ++posi;
    sum += abs(num);
  }
  best = sum;
  if (b[n + n] == -1)
    --nega;
  else
    --posi;
  sum -= abs(a[n + n] - n);
  for (int k = 1; k < n; ++k) {
    int L = n - k + 1, R = L + n - 1;
    sum -= posi, sum += nega;
    b[L] = 1;
    sum += a[L] - 1, ++posi;
    if (sum < best) best = sum, ans = k;
    for (vector<int>::iterator it = c[k].begin(); it != c[k].end(); ++it) {
      int pos = *it;
      if (pos >= L && pos <= R) {
        --posi, ++nega;
        b[pos] = -1;
      }
    }
    if (b[R] == -1)
      --nega;
    else
      --posi;
    sum -= abs(a[R] - n);
  }
  printf("%lld %lld\n", best, ans);
}
int main() {
  input(), prelude(), solve();
  return 0;
}
