#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000100;
int cnt_neg = 0, cnt_pos = 0;
int arr[maxn << 1];
vector<int> faster[maxn * 2];
bool out[maxn * 2];
int main() {
  int n;
  scanf("%d", &n);
  for (int e = 1; e <= n; e++) scanf("%d", arr + e), arr[e + n] = arr[e];
  int hi = 2 * n, lo = n + 1, nowlo = n;
  long long sum = 0;
  for (int cur = hi; cur >= lo; cur--) {
    if (arr[cur] <= nowlo)
      cnt_pos++;
    else
      cnt_neg++, faster[arr[cur] - nowlo].push_back(cur);
    sum += abs(arr[cur] - nowlo);
    nowlo--;
  }
  long long ans = sum, shift = 0, state = 0;
  for (; lo > 1;) {
    if (arr[hi] <= n)
      cnt_pos--;
    else
      cnt_neg--;
    sum -= abs(arr[hi] - n);
    out[hi] = 1;
    shift++;
    sum += -cnt_neg + cnt_pos;
    sum += abs(arr[lo - 1] - 1);
    if (sum < ans) ans = sum, state = shift;
    if (arr[lo - 1] > 1)
      cnt_neg++, faster[shift + arr[lo - 1] - 1].push_back(lo - 1);
    else
      cnt_pos++;
    while (!faster[shift].empty()) {
      if (!out[faster[shift].back()]) cnt_neg--, cnt_pos++;
      faster[shift].pop_back();
    }
    lo--;
    hi--;
  }
  printf("%lld %lld\n", ans, state);
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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 maxn = 1000100;
int cnt_neg = 0, cnt_pos = 0;
int arr[maxn << 1];
vector<int> faster[maxn * 2];
bool out[maxn * 2];
int main() {
  int n;
  scanf("%d", &n);
  for (int e = 1; e <= n; e++) scanf("%d", arr + e), arr[e + n] = arr[e];
  int hi = 2 * n, lo = n + 1, nowlo = n;
  long long sum = 0;
  for (int cur = hi; cur >= lo; cur--) {
    if (arr[cur] <= nowlo)
      cnt_pos++;
    else
      cnt_neg++, faster[arr[cur] - nowlo].push_back(cur);
    sum += abs(arr[cur] - nowlo);
    nowlo--;
  }
  long long ans = sum, shift = 0, state = 0;
  for (; lo > 1;) {
    if (arr[hi] <= n)
      cnt_pos--;
    else
      cnt_neg--;
    sum -= abs(arr[hi] - n);
    out[hi] = 1;
    shift++;
    sum += -cnt_neg + cnt_pos;
    sum += abs(arr[lo - 1] - 1);
    if (sum < ans) ans = sum, state = shift;
    if (arr[lo - 1] > 1)
      cnt_neg++, faster[shift + arr[lo - 1] - 1].push_back(lo - 1);
    else
      cnt_pos++;
    while (!faster[shift].empty()) {
      if (!out[faster[shift].back()]) cnt_neg--, cnt_pos++;
      faster[shift].pop_back();
    }
    lo--;
    hi--;
  }
  printf("%lld %lld\n", ans, state);
  return 0;
}

```