#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
long long dp[17][100010];
long long coef[100010];
long long Calc(int l, int r) {
  long long sum = 0;
  for (int i = l; i <= r;) {
    int v = r / i, j = r / v;
    sum += coef[v] * (j - i + 1ll);
    i = j + 1;
  }
  return sum;
}
void Solve(long long *f, long long *g, int l, int r, int L, int R) {
  if (l > r) return;
  int mid = l + r >> 1, pos = -1;
  if (min(mid - 1, R) >= L) {
    long long wei = Calc(min(mid - 1, R) + 1, mid);
    for (int i = min(mid - 1, R); i >= L; i--) {
      if (g[mid] > f[i] + wei) g[mid] = f[i] + wei, pos = i;
      wei += coef[mid / i];
    }
  }
  if (pos == -1) pos = L;
  Solve(f, g, l, mid - 1, L, pos);
  Solve(f, g, mid + 1, r, pos, R);
}
int main() {
  for (int i = 1; i <= N; i++) {
    coef[i] = i;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i + i; j <= N; j += i) {
      coef[j] -= coef[i];
    }
  }
  for (int i = 1; i <= N; i++) {
    coef[i] += coef[i - 1];
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= N; i++) {
    dp[1][i] = i * (i + 1ll) / 2;
  }
  for (int it = 1; it < 16; it++) {
    Solve(dp[it], dp[it + 1], 1, N, 1, N);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k >= 17)
      printf("%d\n", n);
    else
      printf("%lld\n", dp[k][n]);
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
long long dp[17][100010];
long long coef[100010];
long long Calc(int l, int r) {
  long long sum = 0;
  for (int i = l; i <= r;) {
    int v = r / i, j = r / v;
    sum += coef[v] * (j - i + 1ll);
    i = j + 1;
  }
  return sum;
}
void Solve(long long *f, long long *g, int l, int r, int L, int R) {
  if (l > r) return;
  int mid = l + r >> 1, pos = -1;
  if (min(mid - 1, R) >= L) {
    long long wei = Calc(min(mid - 1, R) + 1, mid);
    for (int i = min(mid - 1, R); i >= L; i--) {
      if (g[mid] > f[i] + wei) g[mid] = f[i] + wei, pos = i;
      wei += coef[mid / i];
    }
  }
  if (pos == -1) pos = L;
  Solve(f, g, l, mid - 1, L, pos);
  Solve(f, g, mid + 1, r, pos, R);
}
int main() {
  for (int i = 1; i <= N; i++) {
    coef[i] = i;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i + i; j <= N; j += i) {
      coef[j] -= coef[i];
    }
  }
  for (int i = 1; i <= N; i++) {
    coef[i] += coef[i - 1];
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= N; i++) {
    dp[1][i] = i * (i + 1ll) / 2;
  }
  for (int it = 1; it < 16; it++) {
    Solve(dp[it], dp[it + 1], 1, N, 1, N);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k >= 17)
      printf("%d\n", n);
    else
      printf("%lld\n", dp[k][n]);
  }
  return 0;
}

```