description: For two positive integers l and r (l ≤ r) let c(l, r) denote the number of integer pairs (i, j) such that l ≤ i ≤ j ≤ r and \operatorname{gcd}(i, j) ≥ l. Here, \operatorname{gcd}(i, j) is the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers i and j.
  
  YouKn0wWho has two integers n and k where 1 ≤ k ≤ n. Let f(n, k) denote the minimum of ∑_{i=1}^{k}{c(x_i+1,x_{i+1})} over all integer sequences 0=x_1 < x_2 < … < x_{k} < x_{k+1}=n.
  
  Help YouKn0wWho find f(n, k). 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 3 ⋅ 10^5) — the number of test cases.
  
  The first and only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 10^5). 
  
  Output
  
  For each test case, print a single integer — f(n, k).
  
  Example
  
  Input
  
  
  4
  6 2
  4 4
  3 1
  10 3
  
  
  Output
  
  
  8
  4
  6
  11
  
  Note
  
  In the first test case, YouKn0wWho can select the sequence [0, 2, 6]. So f(6, 2) = c(1, 2) + c(3, 6) = 3 + 5 = 8 which is the minimum possible.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1e5;
  const long long inf = 1ll << 60;
  bool p[N + 5];
  int np[N + 5], ps;
  long long val[N + 5];
  void prep() {
    p[0] = p[1] = 1;
    val[1] = 1;
    for (int i = 2; i <= N; i++) {
      if (!p[i]) val[np[++ps] = i] = i - 1;
      for (int j = 1; i * np[j] <= N; j++) {
        p[i * np[j]] = 1;
        if (i % np[j])
          val[i * np[j]] = val[i] * (np[j] - 1);
        else {
          val[i * np[j]] = val[i] * np[j];
          break;
        }
      }
    }
    for (int i = 2; i <= N; i++) val[i] += val[i - 1];
  }
  int T;
  long long f[20][N + 5];
  long long s[N + 5];
  void solve(int l, int r, int opl, int opr, int k) {
    if (l > r) return;
    int mid = (l + r) >> 1, op;
    if (opr < mid) {
      s[opr + 1] = 0;
      for (int l = opr + 1, r; l <= mid; l = r + 1) {
        r = mid / (mid / l);
        s[opr + 1] += val[mid / l] * (r - l + 1);
      }
    } else
      s[mid] = 1;
    for (int i = min(mid - 1, opr); i >= opl; i--) s[i] = s[i + 1] + val[mid / i];
    f[k][mid] = inf;
    for (int i = opl; i <= min(mid - 1, opr); i++)
      if (f[k - 1][i] + s[i + 1] <= f[k][mid]) {
        f[k][mid] = f[k - 1][i] + s[i + 1];
        op = i;
      }
    solve(l, mid - 1, opl, op, k);
    solve(mid + 1, r, op, opr, k);
  }
  int main() {
    prep();
    for (int i = 1; i <= N; i++) f[1][i] = 1ll * i * (i + 1) / 2;
    for (int i = 2; i <= 17; i++) {
      for (int j = 1; j < 1 << i; j++) f[i][j] = j;
      solve(1 << i, N, 1, N, i);
    }
    scanf("%d", &T);
    while (T--) {
      int n, k;
      scanf("%d %d", &n, &k);
      if (k > 17)
        printf("%d\n", n);
      else
        printf("%lld\n", f[k][n]);
    }
    return 0;
  }
  