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>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
const int N = 1e6 + 6;
long long int arr[3 * N];
long long int rot_b4_zero[3 * N];
pair<long long int, long long int> ans;
long long int poscnt;
long long int curr;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n, i, j, k, dev, idx;
  cin >> n;
  for (int i = (1); i <= (n); i++) {
    cin >> arr[i];
    ans.first += abs(arr[i] - i);
    if (arr[i] > i) {
      rot_b4_zero[arr[i] - i]++;
      poscnt++;
    }
  }
  curr = ans.first;
  for (int k = (1); k <= (n - 1); k++) {
    curr -= abs(arr[n + 1 - k] - n);
    curr += abs(arr[n + 1 - k] - 1);
    curr -= poscnt;
    curr += n - 1 - poscnt;
    poscnt -= rot_b4_zero[k];
    if (curr < ans.first) ans = {curr, k};
    if (arr[n + 1 - k] > 1) {
      poscnt++;
      rot_b4_zero[k + arr[n + 1 - k] - 1]++;
    }
  }
  cout << ans.first << " " << ans.second << "\n";
  return 0;
}
