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 mod = 998244353;
  long long cal[100005];
  long long val[100005];
  vector<int> prime;
  long long dp[25][100005];
  long long cal2(int l, int r) {
    if (l > r) return 0;
    long long tot = 0;
    long long last = r;
    for (int i = 1;;) {
      int Next = (r) / (i + 1);
      if (Next < l) {
        tot += cal[i] * (last - l + 1);
        break;
      }
      tot += cal[i] * (last - Next);
      last = Next;
      i = r / Next;
    }
    return tot;
  }
  void dc(int k, int l, int r, int optl, int optr) {
    if (l > r) return;
    int mid = (l + r) / 2;
    int opt = 0;
    dp[k][mid] = 1e18;
    long long now = cal2(optr + 1, mid);
    for (int i = min(optr, mid); i >= optl; i--) {
      now += cal[mid / i];
      long long val = dp[k - 1][i - 1] + now;
      if (val < dp[k][mid]) {
        dp[k][mid] = val;
        opt = i;
      }
    }
    dc(k, l, mid - 1, optl, opt);
    dc(k, mid + 1, r, opt, optr);
  }
  void solve() {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k >= 20)
      printf("%d\n", n);
    else {
      printf("%lld\n", dp[k][n]);
    }
  }
  int main() {
    for (int i = 1; i <= 100000; i++) val[i] = i;
    for (int i = 2; i <= 100000; i++) {
      if (val[i] == i) {
        for (int j = i; j <= 100000; j += i) val[j] = val[j] / i * (i - 1);
      }
    }
    cal[1] = 1;
    for (int i = 2; i <= 100000; i++) cal[i] = cal[i - 1] + val[i];
    for (int i = 1; i <= 100000; i++) {
      dp[1][i] = i * 1ll * (i + 1) / 2;
    }
    for (int i = 2; i <= 20; i++) {
      dc(i, 1, 100000, 1, 100000);
    }
    int t = 1;
    scanf("%d", &t);
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  