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;
  inline long long read() {
    long long s = 0, w = 1, ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      s = (s << 3) + (s << 1) + ch - 48;
      ch = getchar();
    }
    return s * w;
  }
  const long long maxn = 1e5 + 50, maxk = 17, inf = 1e16;
  long long n, K, T, f[maxk][maxn];
  long long phi[maxn], pri[maxn], tot, notp[maxn], sum[maxn];
  void sieve() {
    notp[0] = notp[1] = 1;
    phi[1] = 1;
    for (long long i = 2; i < maxn; i++) {
      if (notp[i] == 0) {
        pri[++tot] = i;
        phi[i] = i - 1;
      }
      for (long long j = 1; pri[j] * i < maxn; j++) {
        notp[i * pri[j]] = 1;
        if (i % pri[j] == 0) {
          phi[i * pri[j]] = phi[i] * pri[j];
          break;
        }
        phi[i * pri[j]] = phi[i] * phi[pri[j]];
      }
    }
    for (long long i = 1; i < maxn; i++) sum[i] = sum[i - 1] + phi[i];
  }
  long long w(long long i, long long j) {
    long long res = 0;
    for (long long l = i, r = 0; l <= j; l = r + 1) {
      r = j / (j / l);
      r = min(r, j);
      res += (r - l + 1) * sum[j / l];
    }
    return res;
  }
  void solve(long long k, long long l, long long r, long long x, long long y) {
    if (l > r) return;
    long long mid = (l + r) >> 1, c = w(x + 1, mid), res = inf, pos = 0;
    for (long long i = x; i <= y && i <= mid; i++) {
      if (res > f[k - 1][i] + c) res = f[k - 1][i] + c, pos = i;
      c -= sum[mid / (i + 1)];
    }
    f[k][mid] = res;
    solve(k, l, mid - 1, x, pos);
    solve(k, mid + 1, r, pos, y);
  }
  signed main() {
    sieve();
    n = 100000;
    K = 16;
    for (long long i = 1; i <= n; i++) f[1][i] = i * (i + 1) / 2;
    for (long long k = 2; k <= K; k++) solve(k, 1, n, 1, n);
    T = read();
    while (T--) {
      n = read();
      K = read();
      if (K >= 17)
        printf("%lld\n", n);
      else
        printf("%lld\n", f[K][n]);
    }
    return 0;
  }
  