#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int a[maxn], pos[maxn];
set<pair<int, int> > st;
int main() {
  int n;
  cin >> n;
  long long ans = 0;
  int cnt = 0, id = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
    if (i < a[i])
      st.insert(make_pair(a[i] - i, a[i]));
    else
      cnt++;
    ans += abs(i - a[i]);
  }
  long long sum = ans;
  for (int i = 1; i < n; ++i) {
    int last = n - i + 1;
    cnt--;
    sum += abs(1 - a[last]) - abs(n - a[last]) - int(st.size()) + cnt;
    if (sum < ans) ans = sum, id = i;
    st.insert(make_pair(a[last] - 1 + i, a[last]));
    while (int(st.size()) && st.begin()->first - i <= 0)
      st.erase(st.begin()), cnt++;
  }
  cout << ans << " " << id;
  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 maxn = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int a[maxn], pos[maxn];
set<pair<int, int> > st;
int main() {
  int n;
  cin >> n;
  long long ans = 0;
  int cnt = 0, id = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
    if (i < a[i])
      st.insert(make_pair(a[i] - i, a[i]));
    else
      cnt++;
    ans += abs(i - a[i]);
  }
  long long sum = ans;
  for (int i = 1; i < n; ++i) {
    int last = n - i + 1;
    cnt--;
    sum += abs(1 - a[last]) - abs(n - a[last]) - int(st.size()) + cnt;
    if (sum < ans) ans = sum, id = i;
    st.insert(make_pair(a[last] - 1 + i, a[last]));
    while (int(st.size()) && st.begin()->first - i <= 0)
      st.erase(st.begin()), cnt++;
  }
  cout << ans << " " << id;
  return 0;
}

```