#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int T,N,M;
const ll mo=998244353;

const int prime_max = 10101010;
vector<int> prime;
int NP,divp[prime_max];

vector<int> L,R;

void cprime() {
	if(NP) return;
	for(int i=2;i<prime_max;i++) if(divp[i]==0) {
		//M[i]=NP;
		prime.push_back(i); NP++;
		for(ll j=1LL*i*i;j>=i&&j<prime_max;j+=i) if(divp[j]==0) divp[j]=i;
	}
}

const int MAT=2;
struct Mat { ll v[MAT][MAT]; Mat(){ZERO(v);};};

Mat mulmat(Mat& a,Mat& b,int n=MAT) {
	ll mo2=4*mo*mo;
	int x,y,z; Mat r;
	FOR(x,n) FOR(y,n) r.v[x][y]=0;
	FOR(x,n) FOR(z,n) FOR(y,n) {
		r.v[x][y] += a.v[x][z]*b.v[z][y];
		if(r.v[x][y]>mo2) r.v[x][y] -= mo2;
	}
	FOR(x,n) FOR(y,n) r.v[x][y]%=mo;
	return r;
}

Mat powmat(ll p,Mat a,int n=MAT) {
	int i,x,y; Mat r;
	FOR(x,n) FOR(y,n) r.v[x][y]=0;
	FOR(i,n) r.v[i][i]=1;
	while(p) {
		if(p%2) r=mulmat(r,a,n);
		a=mulmat(a,a,n);
		p>>=1;
	}
	return r;
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cprime();
	
	FORR(p,prime) if(p>3) {
		if(divp[p^2]==0) {
			if(p>(p^2)) R.push_back(p);
			else L.push_back(p);
		}
	}
	
	
	
	cin>>T;
	while(T--) {
		cin>>N>>M;
		
		if(N==1) {
			x=lower_bound(ALL(prime),M+1)-prime.begin();
			cout<<x<<endl;
			continue;
		}
		x=lower_bound(ALL(R),M+1)-R.begin();
		Mat A;
		A.v[1][0]=2*x;
		A.v[0][1]=1;
		A.v[1][1]=1;
		A=powmat(N-1,A);
		ll ret=A.v[0][0]+A.v[1][0]+(A.v[0][1]+A.v[1][1])*(2*x);
		cout<<ret%mo<<endl;
		
	}
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}