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;
  template <class T>
  bool uin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  const int maxn = 100000;
  long long phi[maxn + 1];
  long long cp[maxn + 1];
  long long dp[21][maxn + 1];
  int ceil(int a, int b) {
    if (!a) return 0;
    return (a - 1) / b + 1;
  }
  long long f(int L, int R) {
    int k = 1;
    int r = R;
    long long ans = 0;
    while (k * k <= R && r > L) {
      int nr = R / (k + 1);
      uax(nr, L);
      ans += cp[k] * (r - nr);
      r = nr;
      ++k;
    }
    for (int i = (int)(L); i <= (int)(r); ++i) ans += cp[R / i];
    return ans;
  }
  void dnc(int k, int L, int R, int l, int r) {
    if (L >= R) return;
    int M = (L + R) / 2;
    int bestp = -1;
    int ptr = min(M, r);
    long long ff = f(ptr + 1, M);
    while (ptr >= l) {
      if (uin(dp[k][M], dp[k - 1][ptr] + ff)) {
        bestp = ptr;
      }
      if (ptr) ff += cp[M / ptr];
      --ptr;
    }
    dnc(k, L, M, l, bestp);
    dnc(k, M + 1, R, bestp, r);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(10);
    cout << fixed;
    for (int i = 1; i <= (int)(maxn); ++i) phi[i] = i;
    for (int i = 1; i <= (int)(maxn); ++i)
      for (int j = (int)(2); j <= (int)(maxn / i); ++j) phi[i * j] -= phi[i];
    cp[1] = 1;
    for (int i = (int)(2); i <= (int)(maxn); ++i) cp[i] = cp[i - 1] + phi[i];
    for (int j = 0; j < (int)(21); ++j)
      for (int i = 0; i < (int)(maxn + 1); ++i) dp[j][i] = 1e18;
    dp[0][0] = 0;
    for (int k = (int)(1); k <= (int)(20); ++k) {
      dnc(k, 0, maxn + 1, 0, maxn + 1);
    }
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      if (k > 20)
        cout << n << '\n';
      else
        cout << dp[k][n] << '\n';
    }
    return 0;
  }
  