description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  mt19937 ran(time(0));
  long long mod = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  long long Bpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans *= a;
        ans %= mod;
      }
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  const long long MX = 1 << 19;
  long long t1[2 * MX];
  long long t2[2 * MX];
  void upd(long long x) {
    x += MX;
    for (; x > 0; x >>= 1) {
      t1[x]++;
    }
  }
  void upd2(long long l, long long r, long long x) {
    for (l += MX, r += MX; l <= r; l >>= 1, r >>= 1) {
      if (l & 1) {
        t2[l++] -= x;
      }
      if (~r & 1) t2[r--] -= x;
    }
  }
  long long get1(long long l, long long r) {
    long long ans = 0;
    for (l += MX, r += MX; l <= r; l >>= 1, r >>= 1) {
      if (l & 1) ans += t1[l++];
      if (~r & 1) ans += t1[r--];
    }
    return ans;
  }
  long long get2(long long x) {
    x += MX;
    long long ans = 0;
    while (x > 0) {
      ans += t2[x];
      x >>= 1;
    }
    return ans;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    vector<long long> h(MX + 1);
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      ans += sum + i * a[i];
      long long now = a[i];
      while (now < MX) {
        long long v = get1(now, MX - 1);
        upd2(now + (a[i] == now), MX - 1, a[i]);
        ans -= a[i] * v;
        now += a[i];
      }
      ans += get2(a[i]);
      upd(a[i]);
      h[a[i]]++;
      sum += a[i];
      cout << ans << " ";
    }
  }
  