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 rd() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') f = c == '-' ? -1 : 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * f;
}
void wt(long long x) {
  if (x >= 10) wt(x / 10);
  putchar(x % 10 + 48);
}
const int N = 1e6 + 10;
priority_queue<int, vector<int>, greater<int>> q;
int a[N], n, tot, Ans;
long long ans, u;
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) a[i] = rd();
  for (int i = 1; i <= n; i++) {
    if (a[i] >= i)
      tot++;
    else
      q.push(i - a[i]);
    ans += abs(a[i] - i);
  }
  u = ans;
  for (int k = 1; k <= n; k++) {
    tot--;
    u += n - 2 * a[k] + 1;
    u += tot, u -= q.size();
    q.push(n - a[k] + k);
    while (!q.empty()) {
      if (q.top() - k == 0)
        q.pop(), tot++;
      else
        break;
    }
    if (u < ans) ans = u, Ans = n - k;
  }
  wt(ans), putchar(32), wt(Ans);
  return 0;
}
