#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, M = 3e5 + 10;
int n, m;
int a[N];
long long tr1[M], tr2[M];
void add(long long tr[], int u, int first) {
  for (int i = u; i < M; i += i & -i) tr[i] += first;
}
long long get_sum(long long tr[], int u) {
  long long res = 0;
  for (int i = u; i; i -= i & -i) res += tr[i];
  return res;
}
long long get_range(long long tr[], int r, int l) {
  return get_sum(tr, r) - get_sum(tr, l - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  long long ans = 0, suma = 0;
  for (int i = 1; i <= n; i++) {
    ans += suma;
    for (int first = a[i]; first < M; first += a[i])
      ans -= get_range(tr1, M - 1, first) * a[i];
    ans += 1ll * (i - 1) * a[i];
    ans -= get_sum(tr2, a[i]);
    for (int first = a[i]; first < M; first += a[i]) add(tr2, first, a[i]);
    add(tr1, a[i], 1);
    suma += a[i];
    printf("%lld ", ans);
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You have an array a consisting of n distinct positive integers, numbered from 1 to n. Define p_k as $$$p_k = ∑_{1 ≤ i, j ≤ k} a_i mod a_j, where x \bmod y denotes the remainder when x is divided by y. You have to find and print p_1, p_2, \ldots, p_n$$$. 

Input

The first line contains n — the length of the array (2 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n space-separated distinct integers a_1, …, a_n (1 ≤ a_i ≤ 3 ⋅ 10^5, a_i ≠ a_j if i ≠ j). 

Output

Print n integers p_1, p_2, …, p_n. 

Examples

Input


4
6 2 7 3


Output


0 2 12 22


Input


3
3 2 1


Output


0 3 5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, M = 3e5 + 10;
int n, m;
int a[N];
long long tr1[M], tr2[M];
void add(long long tr[], int u, int first) {
  for (int i = u; i < M; i += i & -i) tr[i] += first;
}
long long get_sum(long long tr[], int u) {
  long long res = 0;
  for (int i = u; i; i -= i & -i) res += tr[i];
  return res;
}
long long get_range(long long tr[], int r, int l) {
  return get_sum(tr, r) - get_sum(tr, l - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  long long ans = 0, suma = 0;
  for (int i = 1; i <= n; i++) {
    ans += suma;
    for (int first = a[i]; first < M; first += a[i])
      ans -= get_range(tr1, M - 1, first) * a[i];
    ans += 1ll * (i - 1) * a[i];
    ans -= get_sum(tr2, a[i]);
    for (int first = a[i]; first < M; first += a[i]) add(tr2, first, a[i]);
    add(tr1, a[i], 1);
    suma += a[i];
    printf("%lld ", ans);
  }
  return 0;
}

```