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;
const int MAXN = 2e6 + 100;
long long bestDev, cur;
int cntPos = 0, cntNeg;
int N, A[MAXN], cnt[MAXN], bestIdx;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < (int)N; i++) scanf("%d", &A[i]), A[i]--;
  for (int i = 0; i < (int)N; i++) bestDev += abs(A[i] - i);
  for (int i = 0; i < (int)N; i++)
    if (A[i] - i >= 0) cnt[A[i] - i]++, cntPos++;
  cur = bestDev;
  cntNeg = N - cntPos;
  for (int i = 0; i < (int)N; i++) {
    if (cur < bestDev) {
      bestDev = cur;
      bestIdx = i;
    }
    cnt[A[N - i - 1] + i + 1]++;
    if (A[N - i - 1] - (N - i - 1) >= 0) cnt[A[N - i - 1] - (N - i - 1)]--;
    if (A[N - i - 1] - (N - i - 1) < i) cntPos++, cntNeg--;
    cntNeg += cnt[i];
    cntPos -= cnt[i];
    cur += cntNeg - cntPos + A[N - i - 1] - abs(A[N - i - 1] - (N - i - 1));
    if (A[N - i - 1] > N - i - 1)
      cur += (A[N - i - 1] - (N - i - 1) + 1) - (N - A[N - i - 1] - 1);
    else
      cur -= i - 1;
  }
  printf("%lld %d\n", bestDev, bestIdx);
}
