description: Theofanis decided to visit his uncle's farm. There are s animals and n animal pens on the farm. For utility purpose, animal pens are constructed in one row.
  
  Uncle told Theofanis that a farm is lucky if you can distribute all animals in all pens in such a way that there are no empty pens and there is at least one continuous segment of pens that has exactly k animals in total.
  
  Moreover, a farm is ideal if it's lucky for any distribution without empty pens.
  
  Neither Theofanis nor his uncle knows if their farm is ideal or not. Can you help them to figure it out?
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
  
  The first and only line of each test case contains three integers s, n, and k (1 ≤ s, n, k ≤ 10^{18}; n ≤ s).
  
  Output
  
  For each test case, print YES (case-insensitive), if the farm is ideal, or NO (case-insensitive) otherwise.
  
  Example
  
  Input
  
  
  4
  1 1 1
  1 1 2
  100 50 200
  56220 47258 14497
  
  
  Output
  
  
  YES
  NO
  NO
  YES
  
  Note
  
  For the first and the second test case, the only possible combination is [1] so there always will be a subsegment with 1 animal but not with 2 animals.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long dv2(long long a) {
    if (a & 1) return a / 2 + 1;
    return a / 2;
  }
  void work(long long s, long long n, long long k) {
    if (s == k) return puts("Yes"), void();
    long long lim = 0;
    lim += (s % k + 1) * dv2(s / k + 1);
    lim += (k - s % k - 1) * dv2(s / k);
    puts(n >= lim ? "Yes" : "No");
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      long long s, n, k;
      scanf("%lld%lld%lld", &s, &n, &k);
      work(s, n, k);
    }
    return 0;
  }
  