#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
const long long NR = 1e6 + 10, oo = 2e9;
const long long MOD = 1e9 + 7;
long long n, a[NR];
vector<long long> v[NR];
long long mars[NR];
void add(long long st, long long dr, long long x) {
  if (st > dr) return;
  mars[st] += x;
  mars[dr + 1] += -x;
  if (x == 1)
    v[dr + 1].push_back(-(dr - st + 1));
  else
    v[dr + 1].push_back(dr - st + 1);
}
void add_(long long st, long long dr, long long x) {
  if (st > dr) return;
  v[st].push_back(x);
  v[dr + 1].push_back(-x);
}
signed main() {
  long long i, j;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] < i) {
      add(2, n - i + 1, 1);
      add_(1, n - i + 1, i - a[i]);
      add(n + 3 - i, n + a[i] - i + 1, -1);
      add_(n + 2 - i, n + a[i] - i + 1, a[i] - 1);
      add(n + a[i] + 1 - i + 1, n - i + 1 + i - 1, 1);
      add_(n + a[i] - i + 1, n, 0);
    }
    if (a[i] == i) {
      add(2, n - i + 1, 1);
      add_(1, n - i + 1, 0);
      add(n + 3 - i, n + a[i] - i + 1, -1);
      add_(n + 2 - i, n + a[i] - i + 1, a[i] - 1);
    }
    if (a[i] > i) {
      add(2, a[i] - i + 1, -1);
      add_(1, a[i] - i + 1, a[i] - i);
      add(a[i] - i + 2, n - i + 1, 1);
      add_(a[i] - i + 1, n - i + 1, 0);
      add(n - i + 3, n, -1);
      add_(n + 2 - i, n, a[i] - 1);
    }
  }
  for (j = 1; j <= n + 1; ++j) {
    mars[j] += mars[j - 1];
  }
  for (j = 1; j <= n + 1; ++j) {
    mars[j] += mars[j - 1];
    while (v[j].size()) {
      mars[j] += v[j].back();
      v[j].pop_back();
    }
  }
  long long best = 1e16, sol = 0;
  for (j = 1; j <= n; ++j) {
    if (mars[j] < best) {
      best = mars[j];
      sol = j - 1;
    }
  }
  cout << best << ' ' << sol << '\n';
  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;
using namespace std::chrono;
const long long NR = 1e6 + 10, oo = 2e9;
const long long MOD = 1e9 + 7;
long long n, a[NR];
vector<long long> v[NR];
long long mars[NR];
void add(long long st, long long dr, long long x) {
  if (st > dr) return;
  mars[st] += x;
  mars[dr + 1] += -x;
  if (x == 1)
    v[dr + 1].push_back(-(dr - st + 1));
  else
    v[dr + 1].push_back(dr - st + 1);
}
void add_(long long st, long long dr, long long x) {
  if (st > dr) return;
  v[st].push_back(x);
  v[dr + 1].push_back(-x);
}
signed main() {
  long long i, j;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] < i) {
      add(2, n - i + 1, 1);
      add_(1, n - i + 1, i - a[i]);
      add(n + 3 - i, n + a[i] - i + 1, -1);
      add_(n + 2 - i, n + a[i] - i + 1, a[i] - 1);
      add(n + a[i] + 1 - i + 1, n - i + 1 + i - 1, 1);
      add_(n + a[i] - i + 1, n, 0);
    }
    if (a[i] == i) {
      add(2, n - i + 1, 1);
      add_(1, n - i + 1, 0);
      add(n + 3 - i, n + a[i] - i + 1, -1);
      add_(n + 2 - i, n + a[i] - i + 1, a[i] - 1);
    }
    if (a[i] > i) {
      add(2, a[i] - i + 1, -1);
      add_(1, a[i] - i + 1, a[i] - i);
      add(a[i] - i + 2, n - i + 1, 1);
      add_(a[i] - i + 1, n - i + 1, 0);
      add(n - i + 3, n, -1);
      add_(n + 2 - i, n, a[i] - 1);
    }
  }
  for (j = 1; j <= n + 1; ++j) {
    mars[j] += mars[j - 1];
  }
  for (j = 1; j <= n + 1; ++j) {
    mars[j] += mars[j - 1];
    while (v[j].size()) {
      mars[j] += v[j].back();
      v[j].pop_back();
    }
  }
  long long best = 1e16, sol = 0;
  for (j = 1; j <= n; ++j) {
    if (mars[j] < best) {
      best = mars[j];
      sol = j - 1;
    }
  }
  cout << best << ' ' << sol << '\n';
  return 0;
}

```