#include <bits/stdc++.h>
using namespace std;
const int MN = 4000002;
int N;
vector<int> P;
int offset = 2000001;
struct BIT {
  vector<int> tree;
  void init() { tree = vector<int>(4 * MN, 0); }
  void upd(int idx, int val, int l, int r, int n) {
    if (idx < l || r < idx) return;
    if (l == r) {
      tree[n] += val;
      return;
    }
    int m = (l + r) >> 1;
    upd(idx, val, l, m, 2 * n);
    upd(idx, val, m + 1, r, 2 * n + 1);
    tree[n] = tree[2 * n] + tree[2 * n + 1];
  }
  int quer(int a, int b, int l, int r, int n) {
    if (b < l || r < a) return 0;
    if (a <= l && r <= b) return tree[n];
    int m = (l + r) >> 1;
    int L = quer(a, b, l, m, 2 * n);
    int R = quer(a, b, m + 1, r, 2 * n + 1);
    return L + R;
  }
} bit;
int main() {
  scanf("%d", &N);
  P.resize(N);
  bit.init();
  long long sum = 0;
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
    bit.upd(P[i] - (i + 1) + offset, 1, 0, MN - 1, 1);
    sum += abs(P[i] - (i + 1));
  }
  long long ans = sum;
  int x = 0;
  for (int i = 1; i < N; i++) {
    int t = bit.quer(0, offset, 0, MN - 1, 1);
    sum += t;
    sum -= (N - t);
    offset++;
    sum -= abs(P[N - i] - N - 1);
    sum += abs(P[N - i] - 1);
    bit.upd(offset + P[N - i] - (N + 1), -1, 0, MN - 1, 1);
    bit.upd(offset + P[N - i] - 1, 1, 0, MN - 1, 1);
    if (ans > sum) {
      ans = sum;
      x = i;
    }
  }
  printf("%lld %d", ans, x);
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MN = 4000002;
int N;
vector<int> P;
int offset = 2000001;
struct BIT {
  vector<int> tree;
  void init() { tree = vector<int>(4 * MN, 0); }
  void upd(int idx, int val, int l, int r, int n) {
    if (idx < l || r < idx) return;
    if (l == r) {
      tree[n] += val;
      return;
    }
    int m = (l + r) >> 1;
    upd(idx, val, l, m, 2 * n);
    upd(idx, val, m + 1, r, 2 * n + 1);
    tree[n] = tree[2 * n] + tree[2 * n + 1];
  }
  int quer(int a, int b, int l, int r, int n) {
    if (b < l || r < a) return 0;
    if (a <= l && r <= b) return tree[n];
    int m = (l + r) >> 1;
    int L = quer(a, b, l, m, 2 * n);
    int R = quer(a, b, m + 1, r, 2 * n + 1);
    return L + R;
  }
} bit;
int main() {
  scanf("%d", &N);
  P.resize(N);
  bit.init();
  long long sum = 0;
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
    bit.upd(P[i] - (i + 1) + offset, 1, 0, MN - 1, 1);
    sum += abs(P[i] - (i + 1));
  }
  long long ans = sum;
  int x = 0;
  for (int i = 1; i < N; i++) {
    int t = bit.quer(0, offset, 0, MN - 1, 1);
    sum += t;
    sum -= (N - t);
    offset++;
    sum -= abs(P[N - i] - N - 1);
    sum += abs(P[N - i] - 1);
    bit.upd(offset + P[N - i] - (N + 1), -1, 0, MN - 1, 1);
    bit.upd(offset + P[N - i] - 1, 1, 0, MN - 1, 1);
    if (ans > sum) {
      ans = sum;
      x = i;
    }
  }
  printf("%lld %d", ans, x);
}

```