因為區間可以差分成字首，我們只用考慮≤ab$\le \frac{a}{b}$的最簡真分數ji$\frac{j}{i}$的個數。

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
#define N 10000010
#define up(x,y) (x=(x+(y))%mod)
using namespace std;
const int mod=998244353;
int pri[N/10],num,mu[N],sm[N],i2=(mod+1)/2;
const int R=1000000;
struct ha
{
    int
 
 cnt,hd[R+5],nxt[R];
    ll a[R],b[R];
    void add(ll x,ll y)
    {
        int id=x%R;
        a[++cnt]=x;b[cnt]=y;
        nxt[cnt]=hd[id];hd[id]=cnt;
    }
    ll qry(ll x)
    {
        int id=x%R;
        for(int p=hd[id];p;p=nxt[p])
            if(a[p]==x) return b[p];
        return 0;   
    }
}H;
void getpri(int
 
 n)
{
    memset(mu,0x3f,sizeof(mu));
    mu[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(mu[i]>1) pri[++num]=i,mu[i]=-1;
        for(int j=1;j<=num&&i*pri[j]<=n;j++)
        {
            if(i%pri[j]==0) {mu[i*pri[j]]=0;break;}     
            mu[i*pri[j]]=-mu[i];
        }
    }
    for(int i=1;i<=n;i++)
        sm[i]=(mu[i]+sm[i-1])%mod;
}
ll qm(ll n)
{
    if(n<=10000000) return sm[n];
    ll tmp=H.qry(n);
    if(tmp) return tmp;
    ll re=1;
    for(ll l=2,r;l<=n;l=r+1)
    {
        ll x=n/l;r=n/x;
        up(re,-(r-l+1)%mod*qm(x));
    }
    H.add(n,re);
    return re;
}
ll likegcd(ll n,ll a,ll b,ll c)
{
    ll tn=n%mod;
    ll re=(tn*(tn+1)%mod*i2%mod*(a/b)%mod+(tn+1)*(c/b)%mod)%mod;
    a%=b;c%=b;ll m=(n*a+c)/b;
    if(a==0) return re;
    return (re+tn*m%mod-likegcd(m-1,b,a,b-c-1)+mod)%mod;

}
ll solve(ll n,ll a,ll b)
{
    ll re=0;
    for(ll l=1,r;l<=n;l=r+1)
    {
        ll x=n/l;r=n/x;
        up(re,likegcd(x,a,b,0)*(qm(r)-qm(l-1)+mod)%mod);
    }
    return re;              
}
int main()
{
    int ca,a,b,c,d;
    ll ans,n;
    getpri(10000000);
    scanf("%d",&ca);
    while(ca--)
    {
        scanf("%lld%d%d%d%d",&n,&a,&b,&c,&d);
        ll ans=solve(n,c,d)-solve(n,a,b)+(b<=n);
        ans=(ans%mod+mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}