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 MAXN = 100010, MAXM = 1000010;
  int t, n, k, mu[MAXM], phi[MAXM];
  long long sumphi[MAXM], dp[MAXM][30];
  void solve(int l, int r, int xl, int xr, int k) {
    if (l > r) {
      return;
    }
    int mid = (l + r) >> 1, np = 0;
    long long nw = 0, nv = 1e18;
    for (int l = min(xr, mid - 1) + 2, r; l <= mid; l = r + 1) {
      r = mid / (mid / l);
      nw += sumphi[mid / l] * (r - l + 1);
    }
    for (int i = min(xr, mid - 1); i >= xl; i--) {
      nw += sumphi[mid / (i + 1)];
      if (dp[i][k - 1] + nw < nv) {
        nv = dp[i][k - 1] + nw;
        np = i;
      }
    }
    assert(xl <= np && np <= xr);
    dp[mid][k] = nv;
    solve(l, mid - 1, xl, np, k);
    solve(mid + 1, r, np, xr, k);
    return;
  }
  int main() {
    for (int i = 1; i <= MAXN - 10; i++) {
      mu[i] = 1, phi[i] = i;
      int cpi = i;
      for (int j = 2; j * j <= cpi; j++) {
        if (cpi % j == 0) {
          int cnt = 0;
          while (cpi % j == 0) {
            cpi /= j, cnt++;
          }
          if (cnt > 1) {
            mu[i] = 0;
          } else {
            mu[i] = -mu[i];
          }
          phi[i] = phi[i] / j * (j - 1);
        }
      }
      if (cpi > 1) {
        mu[i] = -mu[i], phi[i] = phi[i] / cpi * (cpi - 1);
      }
      sumphi[i] = sumphi[i - 1] + phi[i];
      dp[i][1] = 1ll * i * (i + 1) / 2;
    }
    for (int i = 2; i <= 20; i++) {
      solve(i, MAXN - 10, i - 1, MAXN - 10, i);
    }
    scanf("%d", &t);
    for (int ii = 1; ii <= t; ii++) {
      scanf("%d%d", &n, &k);
      if (k <= 20) {
        printf("%lld\n", dp[n][k]);
      } else {
        printf("%d\n", n);
      }
    }
    return 0;
  }
  