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 = 1e6 + 5;
int a[maxn];
long long tag[maxn];
long long dif[maxn];
long long sum[maxn];
void applyInc(int pos, int val, int len) {
  if (len <= 0) return;
  tag[pos] += val - 1;
  tag[pos + len] -= val + len - 1;
  dif[pos]++;
  dif[pos + len]--;
}
void applyDec(int pos, int val, int len) {
  if (len <= 0) return;
  tag[pos] += val + 1;
  dif[pos]--;
  tag[pos + len] -= val - len + 1;
  dif[pos + len]++;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] < i) {
      int pos = 0;
      applyInc(pos, i - a[i], n - i + 1);
      pos += n - i + 1;
      applyDec(pos, a[i] - 1, a[i] - 1);
      pos += a[i] - 1;
      pos++;
      applyInc(pos, 1, n - pos);
    }
    if (a[i] == i) {
      int pos = 1;
      applyInc(pos, 1, n - a[i]);
      pos += n - a[i];
      applyDec(pos, a[i] - 1, n - pos);
    }
    if (a[i] > i) {
      int pos = 0;
      applyDec(pos, a[i] - i, a[i] - i);
      pos += a[i] - i;
      pos++;
      applyInc(pos, 1, n - a[i]);
      pos += n - a[i];
      applyDec(pos, a[i] - 1, n - pos);
    }
  }
  long long t = 0;
  for (int i = 0; i < n; i++) {
    t += dif[i];
    sum[i] += tag[i] + sum[i - 1] + t;
  }
  printf("%lld %d\n", *min_element(sum, sum + n),
         min_element(sum, sum + n) - sum);
  return 0;
}
