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>
using namespace std;
using ll = long long;
template <typename T>
bool in_range(int x, int a, int b) {
  return a <= x && x < b;
}
const int MAXN = 1000005;
int N;
int P[MAXN];
map<int, int> pos;
int cntPos, cntNeg, cntZero;
int main() {
  scanf("%d", &N);
  for (int i = 0, _n = N; i < _n; i++) {
    scanf("%d", &P[i]);
    P[i]--;
  }
  for (int i = 0, _n = N; i < _n; i++) {
    if (P[i] == i)
      cntZero++;
    else if (P[i] > i)
      cntPos++, pos[P[i] - i]++;
    else
      cntNeg++;
  }
  ll curDev = 0;
  for (int i = 0, _n = N; i < _n; i++) curDev += abs(i - P[i]);
  int bestId = 0;
  ll bestDev = (ll)N * N;
  for (int id = 0, _n = N; id < _n; id++) {
    if (curDev < bestDev) {
      bestId = id;
      bestDev = curDev;
    }
    int nextId = (N - 1 - id + N) % N;
    int toRemove = P[nextId] - (N - 1);
    if (toRemove == 0) {
      cntZero--;
    } else if (toRemove > 0) {
      cntPos--;
      pos[toRemove + id]--;
    } else {
      cntNeg--;
    }
    curDev -= abs(toRemove);
    curDev += cntZero + cntNeg - cntPos;
    cntNeg += cntZero;
    cntZero = 0;
    while (true) {
      auto it = pos.begin();
      if (it == pos.end()) {
        break;
      }
      if (it->second == 0) {
        pos.erase(it);
        continue;
      }
      if (it->first == 1 + id) {
        cntPos -= it->second;
        cntZero += it->second;
        pos.erase(it);
      }
      break;
    }
    int toAdd = P[nextId];
    if (toAdd == 0) {
      cntZero++;
    } else if (toAdd > 0) {
      cntPos++;
      pos[toAdd + id + 1]++;
    } else {
      cntNeg++;
    }
    curDev += abs(toAdd);
  }
  cout << bestDev << " " << bestId << endl;
}
