#include <bits/stdc++.h>
using namespace std;
const int MAXV = 3e5;
void add(vector<long long> &h, int idx, long long v) {
  for (; idx < (int)h.size(); idx |= (idx + 1)) h[idx] += v;
}
long long get(vector<long long> &h, int idx) {
  long long res = 0;
  if (idx >= (int)h.size()) idx = h.size() - 1;
  for (; idx >= 0; idx = (idx & (idx + 1)) - 1) res += h[idx];
  return res;
}
long long get(vector<long long> &h, int idx1, int idx2) {
  return get(h, idx2) - get(h, idx1);
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  long long res = 0;
  long long p = 0;
  vector<long long> h(MAXV + 1);
  vector<long long> c(MAXV + 1);
  vector<char> have(MAXV + 1);
  for (int i = 0; i < n; ++i) {
    long long res1 = 0;
    long long res2 = 0;
    int v = a[i];
    res2 += get(c, v, MAXV) * v;
    for (int j = 1; j < MAXV + 1; ++j) {
      int aa = v / (j + 1);
      int bb = v / j;
      if (aa + 3 > bb) {
        for (int j = bb; j >= 0; --j)
          if (have[j]) res2 += v % j;
        break;
      } else {
        long long c1 = get(c, aa, bb);
        long long h1 = get(h, aa, bb);
        res2 += v * c1 - h1 * j;
      }
    }
    long long res4 = 0;
    {
      int j = 0;
      for (int z = 0; z < MAXV + 1; z += v) {
        long long c1 = get(c, z - 1, z - 1 + v);
        long long h1 = get(h, z - 1, z - 1 + v);
        res4 += h1 - v * c1 * j;
        j += 1;
      }
    }
    res += res2;
    res += res4;
    add(c, v, 1);
    add(h, v, v);
    have[v] = 1;
    printf("%I64d ", res);
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Generate a cpp solution 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 MAXV = 3e5;
void add(vector<long long> &h, int idx, long long v) {
  for (; idx < (int)h.size(); idx |= (idx + 1)) h[idx] += v;
}
long long get(vector<long long> &h, int idx) {
  long long res = 0;
  if (idx >= (int)h.size()) idx = h.size() - 1;
  for (; idx >= 0; idx = (idx & (idx + 1)) - 1) res += h[idx];
  return res;
}
long long get(vector<long long> &h, int idx1, int idx2) {
  return get(h, idx2) - get(h, idx1);
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  long long res = 0;
  long long p = 0;
  vector<long long> h(MAXV + 1);
  vector<long long> c(MAXV + 1);
  vector<char> have(MAXV + 1);
  for (int i = 0; i < n; ++i) {
    long long res1 = 0;
    long long res2 = 0;
    int v = a[i];
    res2 += get(c, v, MAXV) * v;
    for (int j = 1; j < MAXV + 1; ++j) {
      int aa = v / (j + 1);
      int bb = v / j;
      if (aa + 3 > bb) {
        for (int j = bb; j >= 0; --j)
          if (have[j]) res2 += v % j;
        break;
      } else {
        long long c1 = get(c, aa, bb);
        long long h1 = get(h, aa, bb);
        res2 += v * c1 - h1 * j;
      }
    }
    long long res4 = 0;
    {
      int j = 0;
      for (int z = 0; z < MAXV + 1; z += v) {
        long long c1 = get(c, z - 1, z - 1 + v);
        long long h1 = get(h, z - 1, z - 1 + v);
        res4 += h1 - v * c1 * j;
        j += 1;
      }
    }
    res += res2;
    res += res4;
    add(c, v, 1);
    add(h, v, v);
    have[v] = 1;
    printf("%I64d ", res);
  }
}
int main() {
  solve();
  return 0;
}

```