#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, LOG = 18;
inline int read() {
  int s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
  return s * f;
}
const long long inf = 1e18;
int prime[N], pcnt;
long long phi[N];
bool vis[N];
void init(int n) {
  phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) {
      prime[++pcnt] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= pcnt && prime[j] * i <= n; ++j) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  for (int i = 1; i <= n; ++i) phi[i] += phi[i - 1];
}
inline long long calc(int l, int r) {
  long long ret = 0;
  for (int i; l <= r; l = i + 1) {
    i = min(r / (r / l), r);
    ret += 1ll * phi[r / l] * (i - l + 1);
  }
  return ret;
}
inline long long c(int l, int r) { return phi[r / l]; }
int T, n, k;
long long f[N][LOG], g[N][LOG];
void solve(int l, int r, int L, int R) {
  if (l > r) return;
  int mid = (l + r) >> 1;
  long long sum = calc(R + 1, mid);
  int pos = 0;
  long long mn = inf;
  for (int i = min(mid, R); i >= L; --i) {
    sum += phi[mid / i];
    if (sum + f[i - 1][k - 1] <= mn) {
      mn = sum + f[i - 1][k - 1];
      pos = i;
    }
  }
  f[mid][k] = mn;
  solve(l, mid - 1, L, pos);
  solve(mid + 1, r, pos, R);
}
int main() {
  n = 100000;
  init(n);
  for (int i = 1; i <= n; ++i) g[i][1] = f[i][1] = (1ll * i * (i + 1)) >> 1;
  for (k = 2; k <= 17; ++k) solve(1, n, 1, n);
  T = read();
  while (T--) {
    n = read();
    k = read();
    if (k >= 20 || n < (1 << k)) {
      printf("%d\n", n);
      continue;
    }
    printf("%lld\n", f[n][k]);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 1e5 + 5, LOG = 18;
inline int read() {
  int s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
  return s * f;
}
const long long inf = 1e18;
int prime[N], pcnt;
long long phi[N];
bool vis[N];
void init(int n) {
  phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) {
      prime[++pcnt] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= pcnt && prime[j] * i <= n; ++j) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  for (int i = 1; i <= n; ++i) phi[i] += phi[i - 1];
}
inline long long calc(int l, int r) {
  long long ret = 0;
  for (int i; l <= r; l = i + 1) {
    i = min(r / (r / l), r);
    ret += 1ll * phi[r / l] * (i - l + 1);
  }
  return ret;
}
inline long long c(int l, int r) { return phi[r / l]; }
int T, n, k;
long long f[N][LOG], g[N][LOG];
void solve(int l, int r, int L, int R) {
  if (l > r) return;
  int mid = (l + r) >> 1;
  long long sum = calc(R + 1, mid);
  int pos = 0;
  long long mn = inf;
  for (int i = min(mid, R); i >= L; --i) {
    sum += phi[mid / i];
    if (sum + f[i - 1][k - 1] <= mn) {
      mn = sum + f[i - 1][k - 1];
      pos = i;
    }
  }
  f[mid][k] = mn;
  solve(l, mid - 1, L, pos);
  solve(mid + 1, r, pos, R);
}
int main() {
  n = 100000;
  init(n);
  for (int i = 1; i <= n; ++i) g[i][1] = f[i][1] = (1ll * i * (i + 1)) >> 1;
  for (k = 2; k <= 17; ++k) solve(1, n, 1, n);
  T = read();
  while (T--) {
    n = read();
    k = read();
    if (k >= 20 || n < (1 << k)) {
      printf("%d\n", n);
      continue;
    }
    printf("%lld\n", f[n][k]);
  }
  return 0;
}

```