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>
const long long oo = 2 * 1000 * 1000 * 1000;
const int _cnt = 1000 * 1000;
const int _p = 1000 * 1000 * 1000 + 7;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
using namespace std;
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
int n, a[1000005], p, q = 0, ans0 = 0;
long long ans = 0, ans1 = 0, num[5000005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] -= i;
    num[(a[i] > 0) ? a[i] : 0]++;
    ans += abs(a[i]);
  }
  p = n;
  ans1 = ans;
  for (int i = 1; i <= n - 1; i++) {
    a[p] -= i - 1;
    num[(a[p] > 0) ? a[p] + q : 0]--;
    ans1 += (long long)2 * num[0] - (long long)2 * abs(a[p]);
    a[p] = n - 1 - abs(a[p]);
    num[0] += num[q + 1];
    q++;
    num[(a[p] > 0) ? a[p] + q : 0]++;
    p--;
    if (ans1 < ans) {
      ans = ans1;
      ans0 = i;
    }
  }
  printf("%I64d %d\n", ans, ans0);
  return 0;
}
