d(x)表示x的约数个数，让你求（l,r<=10^12,r-l<=10^6,k<=10^7）
#include<cstdio>
using namespace std;
#define MOD 998244353ll
#define MAXP 1000100
typedef long long ll;
ll x,y;
int T,K;
bool isNotPrime[MAXP+10];
int num_prime,prime[MAXP+10];
void shai()
{
for(long i = 2 ; i < MAXP ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
for(long j = 0 ; j < num_prime && i * prime[j] < MAXP ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j]))
break;
}
}
}
ll b[1000010],a[1000010];
int main(){
scanf("%d",&T);
shai();
for(;T;--T){
scanf("%lld%lld%d",&x,&y,&K);
for(ll i=x;i<=y;++i){
a[i-x+1ll]=i;
b[i-x+1ll]=1;
}
for(int i=0;i<num_prime;++i){
ll t=x/(ll)prime[i]*(ll)prime[i]+(ll)(x%(ll)prime[i]!=0)*(ll)prime[i];
for(ll j=t;j<=y;j+=(ll)prime[i]){
int cnt=0;
while(a[j-x+1ll]%(ll)prime[i]==0){
a[j-x+1ll]/=(ll)prime[i];
++cnt;
}
b[j-x+1ll]=(b[j-x+1ll]*(((ll)cnt*(ll)K%MOD+1ll)%MOD))%MOD;
}
}
ll ans=0;
for(ll i=x;i<=y;++i){
if((a[i-x+1ll]>1ll)){
b[i-x+1ll]=(b[i-x+1ll]*((ll)K+1ll))%MOD;
}
ans=(ans+b[i-x+1ll])%MOD;
}
printf("%lld\n",ans);
}
return 0;
}