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>
int n, neg[2000006], arr[2000006], po, ne;
long long mindis, sum;
int shift;
void update(int p, int val) {
  for (; p <= 2000000; p += (p & -p)) neg[p] += val;
}
int negsum(int p) {
  int ret = 0;
  for (; p; p -= (p & -p)) ret += neg[p];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    --arr[i];
    if (i - arr[i] >= 0) {
      sum += i - arr[i];
      ++po;
    } else {
      sum += arr[i] - i;
      update(arr[i] - i, 1);
      ++ne;
    }
  }
  mindis = sum;
  int pos = n - 1;
  for (int p = 0; p <= n - 2; p++) {
    if (pos - arr[pos] >= 0)
      po--;
    else {
      update(arr[pos] - pos, -1);
      ne--;
    }
    sum += po + 2 * negsum(p) - ne;
    sum += arr[pos] - (n - 1 - arr[pos]);
    update(p + 1 + arr[pos], 1);
    ++ne;
    if (mindis > sum) {
      mindis = sum;
      shift = p + 1;
    }
    --pos;
  }
  printf("%I64d %d\n", mindis, shift);
}
