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;
  int b[500010], pr[500010], sz;
  long long sp[500010], phi[500010];
  void init() {
    phi[1] = 1;
    for (int i = 2; i <= 100000; i++) {
      if (!b[i]) {
        phi[i] = i - 1;
        pr[++sz] = i;
      }
      for (int j = 1; j <= sz && pr[j] * i <= 100000; j++) {
        b[pr[j] * i] = 1;
        phi[pr[j] * i] = phi[pr[j]] * phi[i];
        if (i % pr[j] == 0) {
          int tmp = i, c = pr[j];
          while (tmp % pr[j] == 0) {
            tmp /= pr[j];
            c *= pr[j];
          }
          phi[i * pr[j]] = phi[tmp] * (c - c / pr[j]);
          break;
        }
      }
    }
    for (int i = 1; i <= 100000; i++) sp[i] = sp[i - 1] + 1ll * phi[i];
  }
  long long dp[21][500010];
  void work(int fl, int l, int r, int al, int ar) {
    if (al > ar) return;
    int mid = (al + ar) >> 1, p = l, lim = min(mid, r);
    long long val = 0;
    for (int i = l, j; i <= mid; i = j + 1) {
      j = mid / (mid / i);
      val += 1ll * (j - i + 1) * sp[mid / i];
    }
    dp[fl][mid] = dp[fl - 1][l - 1] + val;
    for (int i = l; i <= lim; i++) {
      val = val - sp[mid / i];
      long long now = dp[fl - 1][i] + val;
      if (now <= dp[fl][mid]) {
        dp[fl][mid] = now;
        p = i;
      }
    }
    work(fl, l, p, al, mid - 1);
    work(fl, p, r, mid + 1, ar);
  }
  long long inf = 1e18;
  void init_dp() {
    init();
    int n = 100000;
    for (int fl = 1; fl < 20; fl++)
      for (int i = 1; i <= n; i++) dp[fl][i] = inf;
    for (int i = 1; i <= n; i++) dp[1][i] = 1ll * i * (i + 1) / 2;
    for (int fl = 2; fl < 20; fl++) work(fl, 1, n, fl, n);
  }
  int main() {
    int T, n, k;
    init_dp();
    scanf("%d", &T);
    while (T--) {
      long long ans = 0;
      scanf("%d%d", &n, &k);
      if (k >= 20) {
        printf("%d\n", n);
        continue;
      }
      printf("%lld\n", dp[k][n]);
    }
  }
  