Segmented Sieve (UVA 10140 Prime Distance)

/***(Segmented Sieve)**(Prime<=1e18,range[l:r]<=1e6)**(UVA 10140 Prime Distance)***
Given l,r (1<=l,r<=1e18, r-l<=1e6). You have to print 2 min & 2 max distance prime number.
Input: 1 17.   Output: 2,3 are closest, 7,11 are most distant. ***/
#include<cstdio>
#include <vector>
using ll=long long int;
using namespace std;
const ll sz=1e6;
ll fl[sz+5];
vector<ll> primes,NewP;

void sieve(){
    for (ll i=3; i*i<=sz; i+=2)
        if (!fl[i])
            for (ll j=i+i; j<=sz; j+=i)
                fl[j]=1;
    primes.push_back(2);
    for (ll i=3; i<=sz; i+=2)
        if (!fl[i]) primes.push_back(i);
}
/*All prime in [l:r]. 1<=l,r<=1e18. r-l<=1e6.*/
void SegSieve(ll l,ll r){
    memset(fl,0,sizeof fl);
    for (ll i=0; primes[i]*primes[i]<=r; i++){
        ll st=l/primes[i];
        if (st*primes[i]<l) st++;
        if (st==1) st++;
        for (ll j=st*primes[i]; j<=r; j+=primes[i])
            fl[j-l]=1;
    }
    for (ll i=l; i<=r; i++)
        if (!fl[i-l]) NewP.push_back(i);
}
int main(){
    sieve();
    ll l,r;
    while(~scanf("%lld %lld",&l,&r)){
        if (l<2) l=2;
        NewP.clear(), SegSieve(l,r);
        if(NewP.size()<2){
            puts("There are no adjacent primes.");
            continue;
        }
        ll mnl=NewP[0],mnr=NewP[1];
        ll mxl=NewP[0],mxr=NewP[1];
        for (ll i=1; i<NewP.size(); i++){
            if (mnr-mnl>NewP[i]-NewP[i-1])
                mnl=NewP[i-1],mnr=NewP[i];
            if (mxr-mxl<NewP[i]-NewP[i-1])
                mxl=NewP[i-1],mxr=NewP[i];
        }
        printf("%lld,%lld are closest, %lld,%lld are most distant.\n",mnl,mnr,mxl,mxr);
    }
}