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>
  const int N = 1e5 + 10;
  inline int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  int phi[N], sum[N], vis[N], p[N];
  inline void init(int n = 1e5) {
    phi[1] = 1;
    for (int i = 2, cnt = 0; i <= n; ++i) {
      if (!vis[i]) p[++cnt] = i, phi[i] = i - 1;
      for (int j = 1; j <= cnt && i * p[j] <= n; ++j) {
        vis[i * p[j]] = 1, phi[i * p[j]] = phi[i] * (p[j] - 1);
        if (i % p[j] == 0) {
          phi[i * p[j]] = p[j] * phi[i];
          break;
        }
      }
    }
    for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + phi[i];
  }
  inline int64_t calc(int l, int r) {
    if (l > r) return 1LL << 60;
    int64_t ans = 0;
    for (int i = l, j; i <= r; i = j + 1) {
      j = r / (r / i);
      ans += sum[r / i] * (j - i + 1);
    }
    return ans;
  }
  int64_t f[21][N];
  inline void solve(int l, int r, int L, int R, int64_t *f, int64_t *g) {
    if (l > r) return;
    int mid = l + r >> 1, p = 0;
    int64_t w = calc(std::min(mid, R + 1), mid);
    f[mid] = 1LL << 60;
    for (int i = std::min(mid - 1, R); i >= L; --i) {
      if (g[i] + w < f[mid]) f[mid] = g[i] + w, p = i;
      if (i) w += sum[mid / i];
    }
    solve(l, mid - 1, L, p, f, g);
    solve(mid + 1, r, p, R, f, g);
  }
  int main() {
    int n = 1e5, k = 16;
    init();
    for (int i = 1; i <= n; ++i) f[1][i] = 1LL * i * (i + 1) / 2;
    for (int i = 2; i <= k; ++i) solve(1, n, 0, n, f[i], f[i - 1]);
    for (int T = read(); T; --T) {
      int n = read(), k = read();
      printf("%lld\n", k >= 18 || (1 << k) > n ? n : f[k][n]);
    }
    return 0;
  }
  