#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
struct fenwick {
  vector<long long> fw;
  int n;
  void kc(int _n) {
    n = _n;
    fw.assign(n + 5, 0);
  }
  void upd(int i, int val) {
    i++;
    assert(i > 0);
    for (; i <= n; i += i & -i) fw[i] += val;
  }
  long long get(int i) {
    i++;
    if (i <= 0) return 0;
    assert(i > 0);
    long long ans = 0;
    for (; i > 0; i -= i & -i) ans += fw[i];
    return ans;
  }
  void upd_range(int l, int r, int val) {
    upd(l, val);
    upd(r + 1, -val);
  }
  long long get(int l, int r) { return get(r) - get(l - 1); }
};
const int N = 3e5 + 5;
const int MAX = 6e5;
int n;
int a[N];
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long ans = 0;
  fenwick fw1, cnt1, fw2, cnt2;
  fw1.kc(MAX);
  cnt1.kc(MAX);
  fw2.kc(MAX);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 3e5; j += a[i]) {
      int l = j, r = j + a[i] - 1;
      ans += fw1.get(l, r) - cnt1.get(l, r) * j;
    }
    fw1.upd(a[i], a[i]);
    cnt1.upd(a[i], 1);
    ans += 1ll * a[i] * (i - 1) + fw2.get(a[i]);
    for (int j = 0; j <= 3e5; j += a[i]) {
      fw2.upd_range(j, j + a[i] - 1, -j);
    }
    cout << ans << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
}

### 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 dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
struct fenwick {
  vector<long long> fw;
  int n;
  void kc(int _n) {
    n = _n;
    fw.assign(n + 5, 0);
  }
  void upd(int i, int val) {
    i++;
    assert(i > 0);
    for (; i <= n; i += i & -i) fw[i] += val;
  }
  long long get(int i) {
    i++;
    if (i <= 0) return 0;
    assert(i > 0);
    long long ans = 0;
    for (; i > 0; i -= i & -i) ans += fw[i];
    return ans;
  }
  void upd_range(int l, int r, int val) {
    upd(l, val);
    upd(r + 1, -val);
  }
  long long get(int l, int r) { return get(r) - get(l - 1); }
};
const int N = 3e5 + 5;
const int MAX = 6e5;
int n;
int a[N];
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long ans = 0;
  fenwick fw1, cnt1, fw2, cnt2;
  fw1.kc(MAX);
  cnt1.kc(MAX);
  fw2.kc(MAX);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 3e5; j += a[i]) {
      int l = j, r = j + a[i] - 1;
      ans += fw1.get(l, r) - cnt1.get(l, r) * j;
    }
    fw1.upd(a[i], a[i]);
    cnt1.upd(a[i], 1);
    ans += 1ll * a[i] * (i - 1) + fw2.get(a[i]);
    for (int j = 0; j <= 3e5; j += a[i]) {
      fw2.upd_range(j, j + a[i] - 1, -j);
    }
    cout << ans << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
}

```