#include<bits/stdc++.h>

#define INF 1e18
#define int long long
#define maxn 505
#define mod 998244353 

using namespace std;

inline int read(){
	int lre=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-'){
			f=-1;
		}
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		lre=(lre<<3)+(lre<<1)+(c-'0');
		c=getchar();
	}
	return lre*f;
} 

const int SUM=25000;

int n,m; 
int sum=0;
int L,R;
int dp[505][25005];//dp[前几个][最大和]
int dp2[505][25005];
//第二维前缀和化 

void show_dp(){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=SUM;++j){
			cout<<dp[i][j]<<" "; 
		}
		cout<<endl;
	}
}

signed main(){
	
	n=read();m=read();
	dp[0][0]=1;
	for(int i=1;i<=SUM;i++){
		dp[0][i]=1;
	}
	for(int i=1;i<=n;i++){
		L=read(),R=read();
	//	cout<<"L="<<L<<" R="<<R<<endl; 
		for(int sum=L;sum<=SUM;sum++){
			if(sum>=R){
				dp[i][sum]+=(dp[i-1][sum-L]-dp[i-1][sum-R-1]+mod)%mod;
			}else{
				dp[i][sum]+=dp[i-1][sum-L]; 
			}
			dp[i][sum]%=mod;
		//	cout<<"dp["<<i<<"]["<<sum<<"]="<<dp[i][sum]<<endl;
		}
		for(int j=1;j<=SUM;j++){
			dp[i][j]+=dp[i][j-1];
		}
	}
//	show_dp();
	for(int i=1;i<=n;i++){
		for(int j=0;j<=SUM;j++){
			dp2[i][j]=dp[i][j];
		}
	}
	memset(dp,0,sizeof(dp));
	for(int i=0;i<=SUM;i++){
		dp[0][i]=1;
	}
	
	for(int i=1;i<=m;i++){
		L=read(),R=read();
	//	cout<<"L="<<L<<" R="<<R<<endl; 
		for(int sum=L;sum<=SUM;sum++){
			if(sum>=R){
				dp[i][sum]+=(dp[i-1][sum-L]-dp[i-1][sum-R-1]+mod)%mod;
			}else{
				dp[i][sum]+=dp[i-1][sum-L]; 
			}
			dp[i][sum]%=mod;
		//	cout<<"dp["<<i<<"]["<<sum<<"]="<<dp[i][sum]<<endl;
		}
		for(int j=1;j<=SUM;j++){
			dp[i][j]+=dp[i][j-1];
		}
	}
	
	for(int i=1;i<=n;i++){
		for(int j=SUM;j>=1;--j){
			dp2[i][j]-=dp2[i][j-1];
			dp2[i][j]=(dp2[i][j]+mod)%mod;
		}
	} 
	
	for(int i=1;i<=m;i++){
		for(int j=SUM;j>=1;--j){
			dp[i][j]-=dp[i][j-1];
			dp[i][j]=(dp[i][j]+mod)%mod;
		}
	} 
	
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
		//	cout<<"i="<<i<<" j="<<j<<endl; 
			int res=0;
			for(int sum=0;sum<=SUM;sum++){
			//	cout<<"sum="<<sum<<" ";
				res+=(dp2[i][sum]*dp[j][sum])%mod;
			//	cout<<"dp1="<<dp2[i][sum]-dp2[i][sum-1]<<" dp2="<<dp[j][sum]-dp[j][sum]<<endl; 
				res%=mod;
			} 
			printf("%lld ",res);
		}
		putchar('\n');
	} 
	
	return 0;
}
/*
Σx[i]=b
n

非负数解的数量是:C(n+b-1,b)
*/