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 + 10;
  int n, T, p[N], pcnt, f[N];
  long long phi[N], dp[20][N];
  inline void init(int n) {
    phi[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!f[i]) p[++pcnt] = i, phi[i] = i - 1;
      for (int j = 1; j <= pcnt && i * p[j] <= n; ++j) {
        int num = i * p[j];
        f[num] = 1;
        if (i % p[j])
          phi[num] = phi[i] * phi[p[j]];
        else {
          phi[num] = phi[i] * p[j];
          break;
        }
      }
    }
    for (int i = 2; i <= n; ++i) phi[i] += phi[i - 1];
  }
  inline long long calc(int l, int r) {
    long long ret = 0;
    for (int x = l, y = 0; x <= r; x = y + 1) {
      y = r / (r / x);
      ret += phi[r / x] * (y - x + 1);
    }
    return ret;
  }
  inline void solve(int x, int l, int r, int ql, int qr) {
    if (l > r) return;
    int mid = (l + r) >> 1, pos = ql;
    long long now = calc(min(qr + 1, mid), mid);
    dp[x][mid] = 1e15;
    for (int i = min(qr, mid - 1); i >= ql; --i) {
      if (dp[x - 1][i] + now < dp[x][mid])
        dp[x][mid] = dp[x - 1][i] + now, pos = i;
      now += phi[mid / i];
    }
    solve(x, l, mid - 1, ql, pos);
    solve(x, mid + 1, r, pos, qr);
  }
  int main() {
    n = 100000;
    init(n);
    for (int i = 1; i <= n; ++i) dp[1][i] = 1ll * i * (i + 1) / 2;
    for (int i = 2; (1 << i) <= n; ++i) solve(i, i, n, 1, n);
    scanf("%d", &T);
    while (T--) {
      int k;
      scanf("%d%d", &n, &k);
      if (k >= 20 || (1 << k) > n) {
        printf("%d\n", n);
        continue;
      } else
        printf("%lld\n", dp[k][n]);
    }
    return 0;
  }
  