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>
  const long long INF = 1e16;
  const int M = 1e5 + 5;
  int n, k, tot;
  bool vis[M];
  int prime[M];
  long long phi[M], f[M][18];
  inline void Init() {
    phi[1] = 1;
    for (int i = 2; i < M; ++i) {
      if (!vis[i]) prime[++tot] = i, phi[i] = i - 1;
      for (int j = 1; j <= tot && prime[j] * i < M; ++j) {
        vis[prime[j] * i] = 1;
        if (i % prime[j] == 0) {
          phi[i * prime[j]] = phi[i] * prime[j];
          break;
        }
        phi[i * prime[j]] = phi[i] * (prime[j] - 1);
      }
    }
    for (int i = 2; i < M; ++i) phi[i] += phi[i - 1];
  }
  inline long long c(int L, int R) {
    if (L > R) return INF;
    long long res = 0;
    for (int l = L, r; l <= R; l = r + 1) {
      r = R / (R / l);
      res += 1ll * phi[R / l] * (r - l + 1);
    }
    return res;
  }
  inline void Solve(int t, int l, int r, int ql, int qr) {
    if (ql > qr || l > r) return;
    int mid = ql + qr >> 1, pos = l;
    long long val = c(std ::min(r, mid) + 1, mid);
    for (int i = std ::min(r, mid); i >= l; --i) {
      if (f[mid][t] > f[i][t - 1] + val) f[mid][t] = f[i][t - 1] + val, pos = i;
      if (val == INF)
        val = phi[mid / i];
      else
        val += phi[mid / i];
    }
    Solve(t, l, pos, ql, mid - 1);
    Solve(t, pos, r, mid + 1, qr);
  }
  inline int read() {
    int f = 1, s = 0;
    char ch = getchar();
    while (!isdigit(ch)) (ch == '-') && (f = -1), ch = getchar();
    while (isdigit(ch)) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return f * s;
  }
  int main() {
    Init();
    for (int i = 0; i < 18; ++i)
      for (int j = 0; j < M; ++j) f[j][i] = INF;
    f[0][0] = 0;
    for (int i = 1; i < M; ++i) f[i][1] = 1ll * i * (i + 1) / 2;
    for (int i = 2; i < 18; ++i) Solve(i, 1, M - 5, 1, M - 5);
    int t = read();
    while (t--) {
      int n = read(), k = read();
      if (k >= 18 || (1 << k) > n) {
        std ::cout << n << '\n';
        continue;
      }
      std ::cout << f[n][k] << '\n';
    }
    return 0;
  }
  