#include <bits/stdc++.h>
using namespace std;
const int N = (int)4e6;
const long long INF = (long long)1e18;
int n, a[N], c[N];
vector<int> g[N];
int eval(int pos, int jump) { return a[pos] - jump; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    a[i + n] = a[i];
  }
  int cpos = 0;
  int cneg = 0;
  long long pos = 0;
  long long neg = 0;
  for (int i = 0; i < n; i++) {
    c[i] = a[i] - i;
    if (c[i] >= 0) {
      cpos++;
      pos += c[i];
    } else {
      cneg++;
      neg += -c[i];
    }
  }
  for (int i = n; i < n + n; i++) {
    c[i] = a[i] - (n - 1);
  }
  for (int i = 0; i < n + n; i++) {
    if (c[i] < 0) {
      int at = i - c[i];
      int st = max(0, i - (n - 1));
      int d = at - i;
      g[st + d].push_back(i);
    }
  }
  int where = -1;
  long long res = INF;
  for (int i = 0; i < n; i++) {
    if (pos + neg < res) {
      res = pos + neg;
      where = (n - i) % n;
    }
    res = min(res, pos + neg);
    cpos--;
    pos -= eval(i, 0);
    neg -= cneg;
    pos += cpos;
    for (int j = 0; j < g[i + 1].size(); j++) {
      int at = g[i + 1][j];
      cneg--;
      cpos++;
    }
    if (eval(i + n, n - 1) >= 0) {
      cpos++;
      pos += eval(i + n, n - 1);
    } else {
      cneg++;
      neg += -eval(i + n, n - 1);
    }
  }
  printf("%lld %d\n", res, where);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N = (int)4e6;
const long long INF = (long long)1e18;
int n, a[N], c[N];
vector<int> g[N];
int eval(int pos, int jump) { return a[pos] - jump; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    a[i + n] = a[i];
  }
  int cpos = 0;
  int cneg = 0;
  long long pos = 0;
  long long neg = 0;
  for (int i = 0; i < n; i++) {
    c[i] = a[i] - i;
    if (c[i] >= 0) {
      cpos++;
      pos += c[i];
    } else {
      cneg++;
      neg += -c[i];
    }
  }
  for (int i = n; i < n + n; i++) {
    c[i] = a[i] - (n - 1);
  }
  for (int i = 0; i < n + n; i++) {
    if (c[i] < 0) {
      int at = i - c[i];
      int st = max(0, i - (n - 1));
      int d = at - i;
      g[st + d].push_back(i);
    }
  }
  int where = -1;
  long long res = INF;
  for (int i = 0; i < n; i++) {
    if (pos + neg < res) {
      res = pos + neg;
      where = (n - i) % n;
    }
    res = min(res, pos + neg);
    cpos--;
    pos -= eval(i, 0);
    neg -= cneg;
    pos += cpos;
    for (int j = 0; j < g[i + 1].size(); j++) {
      int at = g[i + 1][j];
      cneg--;
      cpos++;
    }
    if (eval(i + n, n - 1) >= 0) {
      cpos++;
      pos += eval(i + n, n - 1);
    } else {
      cneg++;
      neg += -eval(i + n, n - 1);
    }
  }
  printf("%lld %d\n", res, where);
  return 0;
}

```