#include <bits/stdc++.h>
using namespace std;
inline int read(void) {
  register int x = 0, f = 1, ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1000005;
int t[maxn], p[maxn];
int main(void) {
  int n = read(), x = 0, y = 0, ansk = 0;
  long long now = 0, ans = 1ll << 60;
  for (register int i = 1; i <= n; ++i) {
    p[i] = read();
    now += abs(i - p[i]);
    if (i - p[i] < 0)
      ++t[p[i] - i], ++y;
    else
      ++x, ++t[n - i + p[i]];
  }
  ans = now;
  for (register int k = 1; k <= n - 1; ++k) {
    int i = n - k + 1;
    now += x - y;
    now += abs(1 - p[i]) - abs(n + 1 - p[i]);
    x += t[k] - 1;
    y -= t[k] - 1;
    if (ans > now) {
      ans = now;
      ansk = k;
    }
  }
  printf("%lld %d\n", ans, ansk);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
inline int read(void) {
  register int x = 0, f = 1, ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1000005;
int t[maxn], p[maxn];
int main(void) {
  int n = read(), x = 0, y = 0, ansk = 0;
  long long now = 0, ans = 1ll << 60;
  for (register int i = 1; i <= n; ++i) {
    p[i] = read();
    now += abs(i - p[i]);
    if (i - p[i] < 0)
      ++t[p[i] - i], ++y;
    else
      ++x, ++t[n - i + p[i]];
  }
  ans = now;
  for (register int k = 1; k <= n - 1; ++k) {
    int i = n - k + 1;
    now += x - y;
    now += abs(1 - p[i]) - abs(n + 1 - p[i]);
    x += t[k] - 1;
    y -= t[k] - 1;
    if (ans > now) {
      ans = now;
      ansk = k;
    }
  }
  printf("%lld %d\n", ans, ansk);
  return 0;
}

```