#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6;
int n, p[maxn];
int _cnt[maxn * 5], *cnt;
long long suma, sumb;
int cnta, cntb;
void insert(int x) {
  ++cnt[x];
  if (x > 0) {
    suma += x;
    ++cnta;
  } else {
    sumb -= x;
    ++cntb;
  }
}
void remove(int x) {
  --cnt[x];
  if (x > 0) {
    suma -= x;
    --cnta;
  } else {
    sumb += x;
    --cntb;
  }
}
int main(void) {
  scanf("%d", &n);
  cnt = _cnt + maxn * 2;
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
    --p[i];
    insert(p[i] - i);
  }
  int ans = 0;
  long long best = suma + sumb;
  for (int k = 1; k < n; ++k) {
    remove(p[n - k] - (n - 1));
    suma -= cnta;
    sumb += cntb;
    cnta -= cnt[1];
    cntb += cnt[1];
    ++cnt;
    insert(p[n - k] - 0);
    long long tmp = suma + sumb;
    if (tmp < best) {
      best = tmp;
      ans = k;
    }
  }
  printf("%lld %d\n", best, ans);
  return 0;
}

### Prompt

Your challenge is to write 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 = 1e6;
int n, p[maxn];
int _cnt[maxn * 5], *cnt;
long long suma, sumb;
int cnta, cntb;
void insert(int x) {
  ++cnt[x];
  if (x > 0) {
    suma += x;
    ++cnta;
  } else {
    sumb -= x;
    ++cntb;
  }
}
void remove(int x) {
  --cnt[x];
  if (x > 0) {
    suma -= x;
    --cnta;
  } else {
    sumb += x;
    --cntb;
  }
}
int main(void) {
  scanf("%d", &n);
  cnt = _cnt + maxn * 2;
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
    --p[i];
    insert(p[i] - i);
  }
  int ans = 0;
  long long best = suma + sumb;
  for (int k = 1; k < n; ++k) {
    remove(p[n - k] - (n - 1));
    suma -= cnta;
    sumb += cntb;
    cnta -= cnt[1];
    cntb += cnt[1];
    ++cnt;
    insert(p[n - k] - 0);
    long long tmp = suma + sumb;
    if (tmp < best) {
      best = tmp;
      ans = k;
    }
  }
  printf("%lld %d\n", best, ans);
  return 0;
}

```