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;
#pragma comment(linker, "/STACK:102400000,102400000")
long long mul(long long a, long long b) { return (a * b) % (1000000007); }
long long add(long long a, long long b) { return (a + b) % (1000000007); }
long long sub(long long a, long long b) {
  return ((a - b) % (1000000007) + (1000000007)) % (1000000007);
}
void upd(long long &a, long long b) {
  a = (a % (1000000007) + b % (1000000007)) % (1000000007);
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long a2[1234567], a[1234567];
long long c[2500000] = {};
long long abs2(long long a) { return max(a, -a); }
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) a[i] = read() - i, a2[i] = a[i] + i;
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += abs2(a[i]);
  long long p = 0, lcnt = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] < 0)
      lcnt++;
    else
      c[a[i]]++;
  long long m1 = 0, m2 = ans;
  for (int i = 1; i <= n - 1; i++) {
    int v = a[n - i + 1];
    ans -= abs2(v + n - i + 1 - n);
    if (v < i - 1)
      lcnt--;
    else
      c[v]--;
    v += n;
    lcnt += c[i - 1];
    c[i - 1] = 0;
    ans += lcnt - (n - lcnt - 1);
    if (v <= i - 1)
      lcnt++;
    else
      c[v]++;
    ans += abs2(v - i);
    if (ans < m2) {
      m2 = ans;
      m1 = i;
    }
  }
  cout << m2 << ' ' << m1 << endl;
  return 0;
}
