#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e15;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 2e6 + 5;
const int MAXM = 2e6 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long pw(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void inc(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void dec(long long &a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mul(int x, int y) {
  long long z = 1ll * x * y;
  return z - z / MOD * MOD;
}
int n, a[MAXN], c[MAXN], nxt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  long long sum = 0;
  nxt = 0;
  for (int i = 1; i < n + 1; i++) {
    if (i < a[i]) c[a[i] - i]++, nxt++;
    sum += abs(a[i] - i);
  }
  long long ans = sum;
  int k = 0;
  for (int i = 1; i < n + 1; i++) {
    sum = sum + n - 2 * nxt - 1 + abs(a[n - i + 1] - 1) - abs(a[n - i + 1] - n);
    nxt = nxt - c[i] + (a[n - i + 1] == 1 ? 0 : 1);
    c[a[n - i + 1] - 1 + i]++;
    if (ans > sum) ans = sum, k = i;
  }
  printf("%I64d %d\n", ans, k);
}

### 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;
const long long INF = 1e15;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 2e6 + 5;
const int MAXM = 2e6 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long pw(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void inc(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void dec(long long &a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mul(int x, int y) {
  long long z = 1ll * x * y;
  return z - z / MOD * MOD;
}
int n, a[MAXN], c[MAXN], nxt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  long long sum = 0;
  nxt = 0;
  for (int i = 1; i < n + 1; i++) {
    if (i < a[i]) c[a[i] - i]++, nxt++;
    sum += abs(a[i] - i);
  }
  long long ans = sum;
  int k = 0;
  for (int i = 1; i < n + 1; i++) {
    sum = sum + n - 2 * nxt - 1 + abs(a[n - i + 1] - 1) - abs(a[n - i + 1] - n);
    nxt = nxt - c[i] + (a[n - i + 1] == 1 ? 0 : 1);
    c[a[n - i + 1] - 1 + i]++;
    if (ans > sum) ans = sum, k = i;
  }
  printf("%I64d %d\n", ans, k);
}

```