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 unsigned _mod = 998244353;
unsigned mod;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long qpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x;
    y >>= 1ll;
    x = x * x;
  }
  return ret;
}
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
long long _gcd(long long x, long long y) { return y ? _gcd(y, x % y) : x; }
long long n, a[1000010], b[1000010], mn, ans, p, sum;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    a[i] = read(), sum += abs(a[i] - i);
    b[(a[i] - i + n) % n]++;
    p += (a[i] < i) * 2 - 1;
  }
  mn = sum;
  for (int i = 0; i <= n - 2; ++i) {
    p += 2 * b[i] - 2, sum += abs(a[n - i] - 1) - abs(a[n - i] - n) + p + 1;
    if (sum < mn) mn = sum, ans = i + 1;
  }
  cout << mn << ' ' << ans << '\n';
  return 0;
}
