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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1000000;
int n;
int p[MAXN];
int cnt[MAXN + 1];
void upd(int val, int by) {
  if (val < 0 || val > n) return;
  cnt[val] += by;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &p[i]), --p[i];
  memset(cnt, 0, sizeof(cnt));
  int zero = 0;
  long long sumabs = 0;
  int cntneg = 0, cntzero = 0, cntpos = 0;
  for (int i = (0); i < (n); ++i) {
    int cur = p[i] - i;
    upd(cur, +1);
    if (cur < 0)
      ++cntneg;
    else if (cur == 0)
      ++cntzero;
    else if (cur > 0)
      ++cntpos;
    sumabs += abs(cur);
  }
  long long ret = sumabs;
  int retshifts = 0;
  for (int i = (0); i < (n); ++i) {
    int idx = n - i - 1;
    int oval = p[idx] - idx - zero;
    sumabs -= abs(oval);
    upd(oval + zero, -1);
    if (oval < 0)
      --cntneg;
    else if (oval == 0)
      --cntzero;
    else if (oval > 0)
      --cntpos;
    int nval = oval + n;
    sumabs += abs(nval);
    upd(nval + zero, +1);
    if (nval < 0)
      ++cntneg;
    else if (nval == 0)
      ++cntzero;
    else if (nval > 0)
      ++cntpos;
    int cntone = cnt[zero + 1];
    sumabs += cntneg + cntzero - cntpos;
    cntneg += cntzero, cntzero = cntone, cntpos -= cntone;
    ++zero;
    if (sumabs < ret) ret = sumabs, retshifts = i + 1;
  }
  printf("%lld %d\n", ret, retshifts);
}
int main() {
  run();
  return 0;
}
