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 N = 1e6 + 5;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
int n, p[N], a[3 * N], k;
long long ans, pos, neg, poscnt, negcnt;
int main() {
  read(n);
  for (int i = 0, i_end = n; i < i_end; ++i)
    read(p[i]), ++a[(p[i] -= i + 1) + n];
  for (int i = 0, i_end = n; i < i_end; ++i)
    if (p[i] >= 0)
      pos += p[i], ++poscnt;
    else
      neg -= p[i], ++negcnt;
  ans = pos + neg;
  k = 0;
  for (int i = 0, i_end = n - 1; i < i_end; ++i) {
    int j = (n - 1 - i) % n;
    --a[p[j] + n];
    if (p[j] >= i)
      pos -= p[j] - i, --poscnt;
    else
      neg += p[j] - i, --negcnt;
    if (p[j] + n >= i)
      pos += p[j] + n - i, ++poscnt;
    else
      neg -= p[j] + n - i, ++negcnt;
    ++a[p[j] + n + n];
    poscnt -= a[i + n];
    negcnt += a[i + n];
    pos -= poscnt;
    neg += negcnt;
    if (pos + neg < ans) ans = pos + neg, k = i + 1;
  }
  printf("%lld %d\n", ans, k);
  return 0;
}
