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 N = 1e5 + 9;
  const long long inf = 1e12;
  using ll = long long;
  int phi[N];
  void totient() {
    for (int i = 1; i < N; i++) phi[i] = i;
    for (int i = 2; i < N; i++) {
      if (phi[i] == i) {
        for (int j = i; j < N; j += i) phi[j] -= phi[j] / i;
      }
    }
  }
  ll a[N];
  ll c(int l, int r) {
    if (l > r) return inf;
    ll ans = 0;
    for (int i = l, last; i <= r; i = last + 1) {
      last = r / (r / i);
      int x = 0;
      if (i >= l)
        x = last - i + 1;
      else if (last >= l)
        x = last - l + 1;
      ans += a[r / i] * x;
    }
    return ans;
  }
  ll dp[N][17];
  void yo(int i, int l, int r, int optl, int optr) {
    if (l > r) return;
    int mid = (l + r) / 2;
    dp[mid][i] = inf;
    int opt = optl;
    ll cost = c(min(mid, optr) + 1, mid);
    for (int k = min(mid, optr); k >= optl; k--) {
      ll cc = dp[k][i - 1] + cost;
      if (cc <= dp[mid][i]) {
        dp[mid][i] = cc;
        opt = k;
      }
      if (k <= mid) {
        if (cost == inf)
          cost = a[mid / k];
        else
          cost += a[mid / k];
      }
    }
    yo(i, l, mid - 1, optl, opt);
    yo(i, mid + 1, r, opt, optr);
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    totient();
    for (int i = 1; i < N; i++) {
      a[i] = a[i - 1] + phi[i];
    }
    int n = 100000;
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      dp[i][0] = inf;
    }
    for (int i = 1; i <= n; i++) dp[i][1] = 1LL * i * (i + 1) / 2;
    for (int i = 2; i <= 16; i++) yo(i, 1, n, 1, n);
    int q;
    cin >> q;
    while (q--) {
      int n, k;
      cin >> n >> k;
      if (k >= 20 or (1 << k) > n) {
        cout << n << '\n';
      } else {
        cout << dp[n][k] << '\n';
      }
    }
    return 0;
  }
  