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 LL = long long;
  const LL INF = 1e18;
  using namespace std;
  int phi[100100];
  LL a[100100];
  LL dp[100100][18];
  LL c(int l, int r) {
    if (l > r) return INF;
    LL res = 0;
    for (int i = l; i <= r;) {
      int j = r / (r / i);
      assert(j >= i);
      res += (j - i + 1) * a[r / i];
      i = j + 1;
    }
    return res;
  }
  void dq(int k, int l, int r, int tl, int tr) {
    if (l > r) return;
    int m = l + r >> 1;
    dp[m][k] = INF;
    LL cost = c(tr + 1, m);
    int opt = m;
    for (int p = min(m, tr); p >= tl; p--) {
      LL cur = dp[p][k - 1] + cost;
      if (cur <= dp[m][k]) {
        dp[m][k] = cur;
        opt = p;
      }
      if (cost == INF) {
        cost = a[m / p];
      } else
        cost += a[m / p];
    }
    dq(k, l, m - 1, tl, opt);
    dq(k, m + 1, r, opt, tr);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n = 1e5;
    int k = 17;
    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;
        }
      }
    for (int i = 1; i <= n; i++) {
      a[i] = a[i - 1] + phi[i];
    }
    for (int i = 1; i <= n; i++) {
      dp[i][1] = 1LL * i * (i + 1) / 2;
    }
    for (int j = 2; j <= k; j++) {
      dq(j, 1, n, 1, n);
    }
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      if (k > 17 || (1 << k) - 1 >= n) {
        cout << n << '\n';
        continue;
      }
      cout << dp[n][k] << '\n';
    }
  }
  