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 N, v;
const int MAXN = 1000005;
struct BIT {
  long long ft[MAXN];
  void update(long long k, long long v) {
    for (; k < MAXN; k += k & (-k)) ft[k] += v;
  }
  long long query(long long k) {
    long long sum = 0;
    for (; k; k -= k & (-k)) sum += ft[k];
    return sum;
  }
  void update(int l, int r, long long v) {
    update(l, v);
    update(r + 1, -v);
  }
};
struct LazyBIT {
  BIT bitAdd, bitSub;
  void update(long long l, long long r, long long val) {
    bitAdd.update(l, r, val);
    bitSub.update(l, r, (l - 1) * val);
    bitSub.update(r + 1, (-r + l - 1) * val);
  }
  long long query(long long idx) {
    return idx * bitAdd.query(idx) - bitSub.query(idx);
  }
  long long query(long long l, long long r) { return query(r) - query(l); }
};
BIT bit;
LazyBIT Lbit;
void add(int l, int r, int vl, int vr) {
  if (l > r) return;
  if (vl < vr) {
    bit.update(l, r, vl - 1);
    Lbit.update(l, r, 1);
    bit.update(r + 1, N, -(r - l + 1));
  } else if (vl > vr) {
    bit.update(l, r, vl + 1);
    Lbit.update(l, r, -1);
    bit.update(r + 1, N, +(r - l + 1));
  } else {
    bit.update(l, r, vl);
  }
}
int aux[2];
int main() {
  scanf("%d", &N);
  for (int i = (1); i < (N + 1); i++) {
    scanf("%d", &v);
    aux[0] = N - i + 1;
    aux[1] = (v - i + N) % N + 1;
    if (aux[0] >= aux[1]) {
      add(1, aux[1], abs(v - i), 0);
      add(aux[1] + 1, aux[0], abs(v - (v + 1)), abs(v - N));
      add(aux[0] + 1, N, abs(v - 1), abs(v - (i - 1)));
    } else {
      add(1, aux[0], abs(v - i), abs(v - N));
      add(aux[0] + 1, aux[1], abs(v - 1), 0);
      add(aux[1] + 1, N, abs(v - (v + 1)), abs(v - (i - 1)));
    }
  }
  pair<long long, long long> ans(1e18, -1);
  for (int i = (1); i < (N + 1); i++)
    ans = min(ans,
              pair<long long, long long>(bit.query(i) + Lbit.query(i), i - 1));
  cout << ans.first << " " << ans.second << endl;
}
