#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
int n;
int ta[N], tc[N], ret;
int a[N], b[N];
void update(int i, long long v) {
  for (v -= ta[i], ta[i++] += v; i <= n;
       tc[i - 1] += v, i += ((i) & ((i) ^ ((i)-1))))
    ;
}
int query(int i) {
  for (ret = 0; i; ret += tc[i - 1], i ^= ((i) & ((i) ^ ((i)-1))))
    ;
  return ret;
}
int main() {
  int l = 0;
  int r = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] >= i)
      b[a[i] - i]++, l++;
    else
      r++;
  }
  for (int i = 0; i <= n; i++) update(i, b[i]);
  long long now = 0;
  long long ans1 = 0;
  int ans2 = 0;
  for (int i = 1; i <= n; i++) ans1 += abs(a[i] - i);
  now = ans1;
  for (int i = 1; i <= n - 1; i++) {
    r += b[i - 1];
    l -= b[i - 1];
    int k = query(n) - query(i - 1) + i - 1;
    now += r - l - 1 + 2 * a[n - i + 1] - n - 1;
    r--;
    l++;
    b[i + a[n - i + 1] - 1]++;
    update(i + a[n - i + 1] - 1, ta[i + a[n - i + 1] - 1] + 1);
    if (now < ans1) {
      ans1 = now;
      ans2 = i;
    }
  }
  cout << ans1 << " " << ans2;
  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 N = 2e6 + 10;
int n;
int ta[N], tc[N], ret;
int a[N], b[N];
void update(int i, long long v) {
  for (v -= ta[i], ta[i++] += v; i <= n;
       tc[i - 1] += v, i += ((i) & ((i) ^ ((i)-1))))
    ;
}
int query(int i) {
  for (ret = 0; i; ret += tc[i - 1], i ^= ((i) & ((i) ^ ((i)-1))))
    ;
  return ret;
}
int main() {
  int l = 0;
  int r = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] >= i)
      b[a[i] - i]++, l++;
    else
      r++;
  }
  for (int i = 0; i <= n; i++) update(i, b[i]);
  long long now = 0;
  long long ans1 = 0;
  int ans2 = 0;
  for (int i = 1; i <= n; i++) ans1 += abs(a[i] - i);
  now = ans1;
  for (int i = 1; i <= n - 1; i++) {
    r += b[i - 1];
    l -= b[i - 1];
    int k = query(n) - query(i - 1) + i - 1;
    now += r - l - 1 + 2 * a[n - i + 1] - n - 1;
    r--;
    l++;
    b[i + a[n - i + 1] - 1]++;
    update(i + a[n - i + 1] - 1, ta[i + a[n - i + 1] - 1] + 1);
    if (now < ans1) {
      ans1 = now;
      ans2 = i;
    }
  }
  cout << ans1 << " " << ans2;
  return 0;
}

```