#include<cstdio>
#include<iostream>
using namespace std;
const int o=6e5,MOD=998244353;
inline int fix(int x){return x+(x>>31&MOD);}
inline int qp(int b,int f){int res=1;for(;f;f>>=1,b=b*1ll*b%MOD) if(f&1) res=res*1ll*b%MOD;return res;}
int n,T,q,bt=19,w[o],rev[o],coef[o],z[o];
//coef:(x+1)^t%(x^(2n+4)-1)
inline void init(){
	w[1<<(bt-1)]=1;
	for(int i=(1<<(bt-1))+1,j=qp(3,(MOD-1)/(1<<bt));i<(1<<bt);++i) w[i]=w[i-1]*1ll*j%MOD;
	for(int i=(1<<(bt-1));i--;) w[i]=w[i<<1];
	for(int i=1;i<(1<<bt);++i) rev[i]=((rev[i>>1]>>1)|((i&1)<<(bt-1)));
}
inline void ntt(int*a,int n,bool inv){
	for(int i=1;i<n;++i) if(rev[i]<i) swap(a[i],a[rev[i]]);
	for(int md=1;md<n;md<<=1) for(int i=0;i<n;i+=md<<1) for(int j=0,x,y;j<md;++j)
		x=a[i+j],y=a[i+j+md]*1ll*w[j+md]%MOD,a[i+j]=fix(x+y-MOD),a[i+j+md]=fix(x-y);
	if(inv) for(int i=1,j=n-1;i<j;swap(a[i++],a[j--]));
}
inline void maintain(int*a){
	ntt(a,1<<bt,1);
	for(int i=0,j=qp(1<<bt,MOD-2);i<(1<<bt);++i) a[i]=a[i]*1ll*j%MOD;
	for(int i=(1<<bt)-1,j=2*n+4;i>=j;--i) a[i-j]=fix(a[i-j]+a[i]-MOD),a[i]=0;
}
inline int calc(int x,int y){
	x+=y;y=x-2*y;
	return (coef[(x+T+4*n+8)/2%(2*n+4)]*1ll*coef[(y+T+4*n+8)/2%(2*n+4)]+coef[(x+T+2*n+4)/2%(2*n+4)]*1ll*coef[(y+T+2*n+4)/2%(2*n+4)])%MOD;
}
//翻折容斥，横向和竖向的翻折是独立的，且横向和竖向的翻折分别都是上下交错的
//x'=x+y,y'=x-y是旋转45度，这样两个方向独立
//先上翻再下翻x+=2n+4，即x'+=n+2,y'+=n+2，别的方向同理，发现x'-y'模2n+4不变，x'%(n+2)和y'%(n+1)不变且范围内合法的(x',y')通过合理翻折都存在恰一种方式达到
int main(){
	scanf("%d%d%d",&n,&T,&q);init();
	coef[0]=z[0]=z[1]=1;
	for(int t=T;t;t>>=1,maintain(z)){
		ntt(z,1<<bt,0);
		if(t&1){
			ntt(coef,1<<bt,0);
			for(int i=0;i<(1<<bt);++i) coef[i]=coef[i]*1ll*z[i]%MOD;
			maintain(coef);
		}
		for(int i=0;i<(1<<bt);++i) z[i]=z[i]*1ll*z[i]%MOD;
	}
	for(int X0,Y0,X1,Y1;q--;){
		scanf("%d%d%d%d",&X0,&Y0,&X1,&Y1);
		if((X0+Y0+X1+Y1)%2!=T%2) printf("0\n");
		else printf("%lld\n",(calc(X1-X0,Y1-Y0)+0ll+MOD-calc(n*2+2-X1-X0,Y1-Y0)+MOD-calc(X1-X0,n*2+2-Y1-Y0)+calc(n*2+2-X1-X0,n*2+2-Y1-Y0))%MOD);
	}
	return 0;
}