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;
long long pos = 0;
long long neg = 0;
long long ans1 = 1e18;
long long ans2 = 0;
int a[1000005];
int cnt[2000005];
int tot1 = 0;
int tot2 = 0;
int n;
void update(int i, int p) {
  if (a[i] - i >= 0) {
    if (a[i] - n >= 0) {
      cnt[a[i] - i]--;
      pos -= a[i] - n;
      tot1--;
    } else {
      neg -= n - a[i];
      tot2--;
    }
  } else {
    neg -= n - a[i];
    tot2--;
  }
  tot1 -= cnt[p];
  pos -= tot1;
  neg += cnt[p];
  neg += tot2;
  tot2 += cnt[p];
  cnt[a[i] - 1 + p + 1]++;
  pos += a[i] - 1;
  tot1++;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] - i >= 0) {
      cnt[a[i] - i]++;
      pos += a[i] - i;
      tot1++;
    } else {
      neg += i - a[i];
      tot2++;
    }
  }
  int p = 0;
  if (ans1 > neg + pos) {
    ans1 = neg + pos;
    ans2 = p;
  }
  update(n, p);
  p++;
  for (int i = n - 1; i >= 1; i--, p++) {
    if (ans1 > neg + pos) {
      ans1 = neg + pos;
      ans2 = p;
    }
    update(i, p);
  }
  printf("%lld %lld\n", ans1, ans2);
}
