#include <bits/stdc++.h>
using namespace std;
const int mod = 1E9 + 7;
const double eps = 1e-6;
template <typename T>
T abs(T val) {
  return val > 0 ? val : -val;
}
template <class T>
inline void read(T &ret) {
  char c;
  ret = 0;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') ret = ret * 10 + (c - '0'), c = getchar();
}
template <class T>
inline void out(T x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n;
long long p[2000005];
int mark[2000005];
void solve() {
  long long ori = 0;
  int A = 0, B = 0;
  for (int i = 1; i <= n; i++) {
    ori += abs(p[i] - i);
    if (p[i] >= i)
      mark[p[i] - i]++, A++;
    else
      B++;
  }
  long long now = ori;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int id = n - i;
    A -= mark[i];
    B += mark[i];
    now = now - A + B - abs(p[id] - n - 1) + (p[id] - 1);
    mark[i + p[id]]++, A++, B--;
    if (now < ori) ori = now, ans = i + 1;
  }
  printf("%lld %d\n", ori, ans);
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i]);
    solve();
  }
}

### Prompt

Please create a solution in cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1E9 + 7;
const double eps = 1e-6;
template <typename T>
T abs(T val) {
  return val > 0 ? val : -val;
}
template <class T>
inline void read(T &ret) {
  char c;
  ret = 0;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') ret = ret * 10 + (c - '0'), c = getchar();
}
template <class T>
inline void out(T x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n;
long long p[2000005];
int mark[2000005];
void solve() {
  long long ori = 0;
  int A = 0, B = 0;
  for (int i = 1; i <= n; i++) {
    ori += abs(p[i] - i);
    if (p[i] >= i)
      mark[p[i] - i]++, A++;
    else
      B++;
  }
  long long now = ori;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int id = n - i;
    A -= mark[i];
    B += mark[i];
    now = now - A + B - abs(p[id] - n - 1) + (p[id] - 1);
    mark[i + p[id]]++, A++, B--;
    if (now < ori) ori = now, ans = i + 1;
  }
  printf("%lld %d\n", ori, ans);
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i]);
    solve();
  }
}

```