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 = (int)2e6 + 7;
const int MOD = (int)1e9 + 7;
int n;
int p[N];
long long sum[N];
int cnt[N];
void add(int x) {
  int ind = x + n;
  while (ind < N) {
    sum[ind] += x;
    cnt[ind]++;
    ind |= (ind + 1);
  }
}
long long getSum(int pos) {
  int ind = pos + n;
  long long res = 0;
  while (ind >= 0) {
    res += sum[ind];
    ind &= (ind + 1);
    ind--;
  }
  return res;
}
int getCnt(int pos) {
  int ind = pos + n;
  int res = 0;
  while (ind >= 0) {
    res += cnt[ind];
    ind &= (ind + 1);
    ind--;
  }
  return res;
}
long long calc(int pivot) {
  long long cs = pivot * 1LL * getCnt(pivot) - getSum(pivot);
  cs += getSum(n) - getSum(pivot) - pivot * 1LL * (getCnt(n) - getCnt(pivot));
  return cs;
}
long long prefix[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) {
    p[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    add(p[i]);
    int pivot = n - i;
    prefix[i] = calc(pivot);
  }
  memset(sum, 0, sizeof sum);
  memset(cnt, 0, sizeof cnt);
  long long ans = (long long)1e18;
  int pr = -1;
  for (int i = n; i > 0; i--) {
    add(p[i]);
    int pivot = n - i + 1;
    long long cur = calc(pivot - n);
    cur += prefix[i - 1];
    if (cur < ans) {
      ans = cur;
      pr = n - i + 1;
    }
  }
  if (pr == n) pr = 0;
  cout << ans << ' ' << pr;
  return 0;
}
