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
#include <bits/stdc++.h>
using namespace std;
namespace atcoder {
namespace internal {
int ceil_pow2(int n) {
  int x = 0;
  while ((1U << x) < (unsigned int)(n)) x++;
  return x;
}
int bsf(unsigned int n) { return __builtin_ctz(n); }
}  // namespace internal
}  // namespace atcoder
namespace atcoder {
template <class S, S (*op)(S, S), S (*e)()>
struct segtree {
 public:
  segtree() : segtree(0) {}
  segtree(int n) : segtree(std::vector<S>(n, e())) {}
  segtree(const std::vector<S>& v) : _n(int(v.size())) {
    log = internal::ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  S get(int p) {
    assert(0 <= p && p < _n);
    return d[p + size];
  }
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    S sml = e(), smr = e();
    l += size;
    r += size;
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }
    return op(sml, smr);
  }
  S all_prod() { return d[1]; }
  template <bool (*f)(S)>
  int max_right(int l) {
    return max_right(l, [](S x) { return f(x); });
  }
  template <class F>
  int max_right(int l, F f) {
    assert(0 <= l && l <= _n);
    assert(f(e()));
    if (l == _n) return _n;
    l += size;
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!f(op(sm, d[l]))) {
        while (l < size) {
          l = (2 * l);
          if (f(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }
  template <bool (*f)(S)>
  int min_left(int r) {
    return min_left(r, [](S x) { return f(x); });
  }
  template <class F>
  int min_left(int r, F f) {
    assert(0 <= r && r <= _n);
    assert(f(e()));
    if (r == 0) return 0;
    r += size;
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!f(op(d[r], sm))) {
        while (r < size) {
          r = (2 * r + 1);
          if (f(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  std::vector<S> d;
  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
}  // namespace atcoder
using namespace atcoder;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct S {
  long long cnt, sum;
};
S op(S a, S b) { return S{a.cnt + b.cnt, a.sum + b.sum}; }
S e() { return S{0, 0}; }
const int M = 3 * 100000 + 1;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  segtree<S, op, e> seg(M);
  vector<long long> res(n);
  for (int i = (0); i < (n); ++i) {
    if (i) res[i] = res[i - 1];
    res[i] += seg.prod(1, 2).sum;
    if (a[i] >= 2) {
      S q = seg.prod(2, a[i]);
      res[i] += (q.cnt + q.sum);
    }
    for (int j = a[i], k = 0; j < M; j += a[i], ++k) {
      S q = seg.prod(j, min(j + a[i], M));
      long long czlon = (q.sum - q.cnt * k * a[i]);
      if (czlon > 0) {
      }
      res[i] += czlon;
    }
    seg.set(a[i], S{1, a[i]});
  }
  for (int i = (0); i < (n); ++i) {
    cout << res[i] << ' ';
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}
