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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  void debug() { cerr << "\n"; }
  template <typename H, typename... T>
  void debug(H h, T... t) {
    cerr << h;
    if (sizeof...(t)) cerr << ", ";
    debug(t...);
  }
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  const int N = 1e5 + 5, K = 18, mod = 998244353, INF = 1e9 + 5;
  long long dp[K][N];
  int sito[N], phi[N];
  long long sum[N];
  long long Cost(int l, int r) {
    long long ans = 0;
    while (r >= l * 1ll * l) {
      ans += sum[r / l];
      l++;
    }
    int k = r / l;
    for (; k >= 1; k--) {
      int m = r / k;
      ans += (m - l + 1) * sum[k];
      l = m + 1;
    }
    return ans;
  }
  int n;
  long long cost2(int l, int r) { return Cost(n + 1 - r, n + 1 - l); }
  void prop(int l, int r, int a, int b, int k) {
    int m = (l + r) / 2, skad = m;
    dp[k][m] = dp[k - 1][m];
    long long C = Cost(a, m);
    for (int i = a; i <= b && i < m; i++) {
      C -= sum[m / i];
      if (dp[k - 1][i] + C < dp[k][m]) {
        skad = i;
        dp[k][m] = dp[k - 1][i] + C;
      }
    }
    if (m > l) prop(l, m - 1, a, skad, k);
    if (m < r) prop(m + 1, r, skad, b, k);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    sum[1] = 1;
    for (int i = 2; i < N; i++) {
      if (sito[i] == 0) {
        for (int j = 1; j * i < N; j++) {
          sito[j * i] = i;
        }
      }
      phi[i] = 1;
      int k = i;
      while (k > 1) {
        int p = sito[k];
        phi[i] *= p - 1;
        k /= p;
        while (k % p == 0) {
          k /= p;
          phi[i] *= p;
        }
      }
      sum[i] = sum[i - 1] + phi[i];
    }
    for (int i = 1; i < N; i++) {
      dp[0][i] = Cost(1, i);
    }
    for (int i = 1; i < K; i++) {
      prop(1, N - 1, 1, N - 1, i);
    }
    int t;
    cin >> t;
    while (t--) {
      int k;
      cin >> n >> k;
      if (k >= K || (1 << k) > n)
        cout << n << "\n";
      else {
        cout << dp[k - 1][n] << "\n";
      }
    }
  }
  