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;
int a[N];
long long sum[N], d[N], ans[N];
int n;
void add(int L, int R, int b, int k) {
  if (L > R) {
    return;
  }
  sum[L] += b;
  sum[R + 1] -= b;
  sum[R + 1] -= (R - L + 1) * 1LL * k;
  d[L] += k;
  d[R + 1] -= k;
}
void calc() {
  long long curdf = 0;
  for (int i = 0; i <= n; i++) {
    curdf += d[i];
    sum[i] += curdf;
  }
  long long cursm = 0;
  for (int i = 0; i <= n; i++) {
    cursm += sum[i];
    ans[i] += cursm;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] >= i) {
      add(0, a[i] - i, a[i] - i + 1, -1);
      add(a[i] - i + 1, n - i, 0, 1);
      add(n - i + 1, n - 1, a[i], -1);
    } else {
      add(0, n - i, i - a[i] - 1, 1);
      add(n - i + 1, n - i + a[i], a[i], -1);
      add(n - i + a[i] + 1, n - 1, 0, 1);
    }
  }
  calc();
  long long best = ans[0], besti = 0;
  for (int i = 1; i < n; i++) {
    if (ans[i] < best) {
      best = ans[i];
      besti = i;
    }
  }
  cout << best << " " << besti << '\n';
  return 0;
}
