分析：

要是在一年前，我估计会分类到莫比乌斯反演。。。

很显然的莫比乌斯函数来容斥，只不过算 $10^9$级别的前缀和需要用杜教筛

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 5000010
#define MAXK 1010
#define MOD 1000000007
using namespace std;
typedef long long ll;
int n,k;
vector<int>primes;
bool isprime[MAXN];
int mu[MAXN],tot,N;
map<int,int> used;
void prepare(){
	mu[1]=1;
	N=5000000;
	for(int i=2;i<=N;i++){
		if(isprime[i]==0){
			mu[i]=-1;
			primes.push_back(i);	
		}
		for(int j=0;1ll*i*primes[j]<=1ll*N;j++){
			isprime[i*primes[j]]=1;
			if(i%primes[j]==0)
				break;
			mu[i*primes[j]]=-mu[i];
		}
	}
	for(int i=1;i<=N;i++)
		mu[i]=(mu[i-1]+mu[i]+MOD)%MOD;
}
int fsp(int x,int y){
	int res=1;
	while(y){
		if(y&1)
			res=1ll*res*x%MOD;
		x=1ll*x*x%MOD;
		y>>=1;
	}
	return res;
}
int fac[MAXN],ifac[MAXN];
int pmu(ll x){
	if(x<=N)
		return mu[x];
	if(used.count(x))
		return used[x];
	int res=1;
	int las=1;
	for(int i=2;i<=x;i=las+1){
		las=x/(x/i);
		res=1ll*(res-1ll*pmu(x/i)*(las-i+1)%MOD+MOD)%MOD;
	}
	used[x]=res;
	return res;
}
int C(int n,int m){
	if(n<=N)
		return 1ll*fac[n]*ifac[m]%MOD*ifac[n-m]%MOD;
	int res=ifac[m];
	for(int i=n;i>n-m;i--)
		res=1ll*res*i%MOD;
//	PF("[%d %d %lld]\n",n,m,res);
	return res;
}
int main(){
	freopen("b.in","r",stdin);
	freopen("b.out","w",stdout);
	prepare();
	int T;
	SF("%d",&T);	
	while(T--){
		SF("%d%d",&n,&k);
		fac[0]=1;
		for(int i=1;i<=N;i++)
			fac[i]=1ll*fac[i-1]*i%MOD;
		ifac[N]=fsp(fac[N],MOD-2);
		for(int i=N;i>=1;i--)
			ifac[i-1]=1ll*ifac[i]*i%MOD;
//		PF("[%lld]",fac[k]);
		int las=1;
		ll ans=0;
		for(int i=1;i<=n;i=las+1){
			las=n/(n/i);
			(ans+=1ll*(pmu(las)-pmu(i-1)+MOD)%MOD*C(n/i+k-1,k)%MOD)%=MOD;	
		}
		PF("%lld\n",ans);
	}
}