#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
const int MAX_N = 1e6 + 100;
int N, Nr[MAX_N];
int main() {
  cin >> N;
  for (int(i) = 0; (i) < (int)(N); (i)++) scanf("%d", &Nr[i]), Nr[i]--;
  long long ans = 0, now = 0;
  int ix = 0;
  deque<int> hi(N, 0), lo(N, 0);
  int allH = 0, allL = 0;
  for (int(i) = 0; (i) < (int)(N); (i)++) {
    if (Nr[i] >= i)
      hi[Nr[i] - i]++, allH++;
    else
      lo[i - Nr[i]]++, allL++;
    now += abs(Nr[i] - i);
  }
  ans = now;
  ix = 0;
  for (int it = 1; it < N; it++) {
    int v = Nr[N - it];
    now -= N - 1 - v;
    lo[N - 1 - v]--;
    allL--;
    int f = hi[0];
    allH -= f;
    hi.pop_front();
    hi.push_back(0);
    lo[0] += f;
    allL += f;
    lo.push_front(0);
    lo.pop_back();
    now -= allH;
    now += allL;
    now += v - 0;
    hi[v - 0]++;
    allH++;
    if (ans > now) {
      ans = now;
      ix = it;
    }
  }
  printf("%lld %d\n", ans, ix);
  return 0;
}

### 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 INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
const int MAX_N = 1e6 + 100;
int N, Nr[MAX_N];
int main() {
  cin >> N;
  for (int(i) = 0; (i) < (int)(N); (i)++) scanf("%d", &Nr[i]), Nr[i]--;
  long long ans = 0, now = 0;
  int ix = 0;
  deque<int> hi(N, 0), lo(N, 0);
  int allH = 0, allL = 0;
  for (int(i) = 0; (i) < (int)(N); (i)++) {
    if (Nr[i] >= i)
      hi[Nr[i] - i]++, allH++;
    else
      lo[i - Nr[i]]++, allL++;
    now += abs(Nr[i] - i);
  }
  ans = now;
  ix = 0;
  for (int it = 1; it < N; it++) {
    int v = Nr[N - it];
    now -= N - 1 - v;
    lo[N - 1 - v]--;
    allL--;
    int f = hi[0];
    allH -= f;
    hi.pop_front();
    hi.push_back(0);
    lo[0] += f;
    allL += f;
    lo.push_front(0);
    lo.pop_back();
    now -= allH;
    now += allL;
    now += v - 0;
    hi[v - 0]++;
    allH++;
    if (ans > now) {
      ans = now;
      ix = it;
    }
  }
  printf("%lld %d\n", ans, ix);
  return 0;
}

```