#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long pos[N], s1[N], s2[N];
int a[N];
int n;
void add(long long s[], int pos, long long v) {
  while (pos <= n) {
    s[pos] += v;
    pos += pos & (-pos);
  }
}
long long query(long long s[], int pos) {
  long long res = 0;
  while (pos) {
    res += s[pos];
    pos -= pos & (-pos);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += i - 1 - query(s1, pos[i]);
    add(s1, pos[i], 1);
    add(s2, pos[i], pos[i]);
    int l = 1, r = n, mid;
    while (l <= r) {
      mid = (l + r) >> 1;
      if (query(s1, mid) * 2 <= i)
        l = mid + 1;
      else
        r = mid - 1;
    }
    long long ans2 = 0;
    long long cnt = query(s1, mid), sum = query(s2, mid);
    ans2 += mid * cnt - sum - cnt * (cnt - 1) / 2;
    cnt = i - cnt;
    sum = query(s2, n) - sum;
    ans2 += sum - cnt * (mid + 1) - cnt * (cnt - 1) / 2;
    printf("%lld ", ans + ans2);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given a permutation p_1, p_2, …, p_n.

In one move you can swap two adjacent values.

You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.

Let f(k) be the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation.

You need to find f(1), f(2), …, f(n).

Input

The first line of input contains one integer n (1 ≤ n ≤ 200 000): the number of elements in the permutation.

The next line of input contains n integers p_1, p_2, …, p_n: given permutation (1 ≤ p_i ≤ n).

Output

Print n integers, the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation, for k=1, 2, …, n.

Examples

Input


5
5 4 3 2 1


Output


0 1 3 6 10 


Input


3
1 2 3


Output


0 0 0 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long pos[N], s1[N], s2[N];
int a[N];
int n;
void add(long long s[], int pos, long long v) {
  while (pos <= n) {
    s[pos] += v;
    pos += pos & (-pos);
  }
}
long long query(long long s[], int pos) {
  long long res = 0;
  while (pos) {
    res += s[pos];
    pos -= pos & (-pos);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += i - 1 - query(s1, pos[i]);
    add(s1, pos[i], 1);
    add(s2, pos[i], pos[i]);
    int l = 1, r = n, mid;
    while (l <= r) {
      mid = (l + r) >> 1;
      if (query(s1, mid) * 2 <= i)
        l = mid + 1;
      else
        r = mid - 1;
    }
    long long ans2 = 0;
    long long cnt = query(s1, mid), sum = query(s2, mid);
    ans2 += mid * cnt - sum - cnt * (cnt - 1) / 2;
    cnt = i - cnt;
    sum = query(s2, n) - sum;
    ans2 += sum - cnt * (mid + 1) - cnt * (cnt - 1) / 2;
    printf("%lld ", ans + ans2);
  }
  return 0;
}

```