#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, mod = 1e9 + 7;
struct BIT {
  vector<long long> tree;
  BIT(int n) : tree(n + 2) {}
  void upd(int p, long long v) {
    for (int i = ++p; i < tree.size(); i += i & -i) tree[i] += v;
  }
  long long get(int p) {
    long long sum = 0;
    for (int i = ++p; i; i -= i & -i) sum += tree[i];
    return sum;
  }
  long long query(int l, int r) {
    if (r < l) return 0;
    return get(r) - get(l - 1);
  }
  int get_kth(long long k) {
    int pos = 0;
    long long s = 0;
    for (int i = 1 + log2(tree.size()); i >= 0; i--)
      if (pos + (1 << i) < tree.size() && s + tree[pos + (1 << i)] < k)
        s += tree[pos + (1 << i)], pos += (1 << i);
    return pos;
  }
  int lower_bound(int v) { return get_kth(get(v - 1) + 1); }
  int upper_bound(int v) { return lower_bound(v + 1); }
};
void Run() {
  int n;
  cin >> n;
  BIT freq(2 * N), vals(2 * N);
  vector<long long> a(n + 1);
  vector<long long> s(n + 1), t(n + 1);
  long long sum = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    s[i] = s[i - 1] + (i - 1) * a[i] + vals.get(a[i]);
    for (int j = a[i]; j < N; j += a[i]) {
      vals.upd(j, -j);
      vals.upd(j + a[i], j);
    }
    t[i] = t[i - 1] + sum;
    for (int j = a[i]; j < N; j += a[i])
      t[i] -= freq.query(j, j + a[i] - 1) * j;
    sum += a[i];
    freq.upd(a[i], 1);
    cout << s[i] + t[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  while (t--) Run();
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 3e5 + 5, mod = 1e9 + 7;
struct BIT {
  vector<long long> tree;
  BIT(int n) : tree(n + 2) {}
  void upd(int p, long long v) {
    for (int i = ++p; i < tree.size(); i += i & -i) tree[i] += v;
  }
  long long get(int p) {
    long long sum = 0;
    for (int i = ++p; i; i -= i & -i) sum += tree[i];
    return sum;
  }
  long long query(int l, int r) {
    if (r < l) return 0;
    return get(r) - get(l - 1);
  }
  int get_kth(long long k) {
    int pos = 0;
    long long s = 0;
    for (int i = 1 + log2(tree.size()); i >= 0; i--)
      if (pos + (1 << i) < tree.size() && s + tree[pos + (1 << i)] < k)
        s += tree[pos + (1 << i)], pos += (1 << i);
    return pos;
  }
  int lower_bound(int v) { return get_kth(get(v - 1) + 1); }
  int upper_bound(int v) { return lower_bound(v + 1); }
};
void Run() {
  int n;
  cin >> n;
  BIT freq(2 * N), vals(2 * N);
  vector<long long> a(n + 1);
  vector<long long> s(n + 1), t(n + 1);
  long long sum = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    s[i] = s[i - 1] + (i - 1) * a[i] + vals.get(a[i]);
    for (int j = a[i]; j < N; j += a[i]) {
      vals.upd(j, -j);
      vals.upd(j + a[i], j);
    }
    t[i] = t[i - 1] + sum;
    for (int j = a[i]; j < N; j += a[i])
      t[i] -= freq.query(j, j + a[i] - 1) * j;
    sum += a[i];
    freq.upd(a[i], 1);
    cout << s[i] + t[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  while (t--) Run();
}

```