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.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long linf = 1e18 + 7;
const int inf = 1e9 + 7;
const int N = 1e6 + 5;
int n, a[N], cnt, pc, mc;
long long ans = linf, ans_p, cur;
deque<int> dq;
set<pair<int, int> > st;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int x = i - a[i];
    if (x >= 0) {
      cur += x;
      pc++;
    } else {
      cur -= x;
      mc++;
      st.insert(make_pair(-x, i));
    }
    dq.push_back(x);
  }
  for (int i = n; i >= 1; i--) {
    if (cur < ans) {
      ans = cur;
      ans_p = n - i;
    }
    int x = dq.back();
    if (x + cnt >= 0) {
      cur -= x + cnt;
      pc--;
    } else {
      cur += x + cnt;
      mc--;
    }
    dq.pop_back();
    cur += pc;
    cur -= mc;
    cnt++;
    x = 1 - a[i] - cnt;
    if (x + cnt >= 0) {
      cur += x + cnt;
      pc++;
    } else {
      cur -= x + cnt;
      mc++;
      st.insert(make_pair(-x, i + n));
    }
    dq.push_front(x);
    while (!st.empty()) {
      pair<int, int> pr = *st.begin();
      pr.first *= -1;
      if (pr.first + cnt >= 0) {
        cur += pr.first + cnt;
        pc++;
        mc--;
        st.erase(st.begin());
      } else
        break;
    }
  }
  cout << ans << " " << ans_p << endl;
}
