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 a[2000005];
int n;
int tp[1000005];
long long ans = 0;
int x = 0;
int neg = 0, pos = 0;
vector<int> modify[2000005];
void solve() {
  for (int i = 1; i <= n; i++) {
    neg += a[i] - i < 0;
    pos += a[i] - i >= 0;
    ans += abs(a[i] - i);
    if (a[i] < i) modify[i - a[i]].push_back(i);
  }
  long long cnt = ans;
  for (int i = 1; i < n; i++) {
    int l = i + 1, r = i + n;
    cnt -= a[i] - 1, cnt += abs(a[r] - n);
    pos--;
    cnt += pos - neg;
    neg += a[r] - n < 0, pos += a[r] - n >= 0;
    for (vector<int>::iterator j = modify[i].begin(); j != modify[i].end();
         ++j) {
      if (*j <= i) continue;
      pos++, neg--;
    }
    if (a[r] < n) modify[n - a[r] + i].push_back(r);
    if (cnt < ans) ans = cnt, x = n - i;
  }
  cout << ans << " " << x << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) a[i + n] = a[i];
  solve();
  return 0;
}
