#include <bits/stdc++.h> 
#include <assert.h> 
#include <unordered_set>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define LL long long
#define pp pair<LL,int>
#define mp make_pair 
#define ull unsigned long long
namespace IO{
    const int sz=1<<22;
    char a[sz+5],b[sz+5],*p1=a,*p2=a,*t=b,p[105];
    inline char gc(){
		return getchar();
    //    return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
    }
    template<class T> void gi(T& x){
        x=0; int f=1;char c=gc();
        if(c=='-')f=-1;
        for(;c<'0'||c>'9';c=gc())if(c=='-')f=-1;
        for(;c>='0'&&c<='9';c=gc())
            x=x*10+(c-'0');
        x=x*f;
    }
    inline void flush(){fwrite(b,1,t-b,stdout),t=b; }
    inline void pc(char x){*t++=x; if(t-b==sz) flush(); }
    template<class T> void pi(T x,char c='\n'){
        if(x<0)pc('-'),x=-x;
        if(x==0) pc('0'); int t=0;
        for(;x;x/=10) p[++t]=x%10+'0';
        for(;t;--t) pc(p[t]); pc(c);
    }
    struct F{~F(){flush();}}f; 
}
using IO::gi;
using IO::pi;
using IO::pc;
using IO::gc;
const int mod=998244353;
const int inv2=(mod+1)>>1;
inline int add(int x,int y){
	return x+y>=mod?x+y-mod:x+y;
}
inline int dec(int x,int y){
	return x-y<0?x-y+mod:x-y;
}
inline int qkpow(int a,int b){
	int ans=1,base=a;
	while(b){
		if(b&1)ans=1ll*ans*base%mod;
		base=1ll*base*base%mod;
		b>>=1;
	}
	return ans;
}
int fac[1000005],inv[1000005],Invn[1000005];
inline int binom(int n,LL m){
	if(n<m||m<0)return 0;
	return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
void init_C(int N){
	fac[0]=1; 
	for(int i=1;i<=N;i++)fac[i]=1ll*fac[i-1]*i%mod; 
	inv[0]=1;
	inv[N]=qkpow(fac[N],mod-2);
	for(int i=N-1;i>=1;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
	Invn[0]=Invn[1]=1;
	for(int i=2;i<=N;i++)Invn[i]=1ll*inv[i]*fac[i-1]%mod;
}     
const int MAXN=6e5+5;
int rev[MAXN],F[MAXN],Q[MAXN],W[23][MAXN<<1],f[MAXN],g[MAXN];
inline int mul(int a,int b){return 1ll*a*b%mod;}
void init(){
    int wn;
    for(int i=0,x=1;x<MAXN;i++,x<<=1){
        W[i][x]=1;
        wn=qkpow(3,(mod-1)/(x<<1));
        for(int j=1;j<x;j++)W[i][x+j]=1ll*wn*W[i][x+j-1]%mod;
        wn=qkpow(wn,mod-2);
        for(int j=1;j<x;j++)W[i][x-j]=1ll*wn*W[i][x-j+1]%mod;
    } 
} 
inline void NTT(int *a,int type,int n){
    for(int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    LL x,y;
    for(int i=1,cnt=0;i<n;cnt++,i<<=1){
        for(int j=0;j<n;j+=(i<<1)){
            for(int t=0,k=j;k<j+i;k++,t+=type){
                x=a[k],y=1ll*W[cnt][i+t]*a[k+i];
                a[k]=(x+y)%mod;
                a[k+i]=(x-y)%mod;
            }
        }
    }
    if(type==-1)for(int i=0;i<n;i++)a[i]=mul(a[i],Invn[n]);
}
void Polyinv(int *a,int *b,int len){
	static int c[MAXN];
	if(len==1){
		b[0]=qkpow(a[0],mod-2);
		return ;
	}
	Polyinv(a,b,(len+1)>>1);
	int l=0,m=(len<<1),n=1;
	for(;n<=m;n<<=1,l++);
	for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
	for(int i=0;i<len;i++)c[i]=a[i];
	for(int i=len;i<n;i++)c[i]=0;
	NTT(c,1,n),NTT(b,1,n);
	for(int i=0;i<n;i++)b[i]=1ll*(2-1ll*b[i]*c[i])%mod*b[i]%mod;
	NTT(b,-1,n);
	for(int i=len;i<n;i++)b[i]=0;
}
inline int init_NTT(int L){
	int l=0,m=L,n=1;
	for(;n<=m;n<<=1,l++);
	for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
	return n;
}
void MUL(int *a,int *b,int *c,int n,int m,int lim){
	static int t1[MAXN],t2[MAXN],t3[MAXN];
	for(int i=0;i<n;i++)t1[i]=a[i];
	for(int i=0;i<m;i++)t2[i]=b[i];
	for(int i=n;i<lim;i++)t1[i]=0;
	for(int i=m;i<lim;i++)t2[i]=0;
	NTT(t1,1,lim),NTT(t2,1,lim);
	for(int i=0;i<lim;i++)t3[i]=1ll*t1[i]*t2[i]%mod;
	NTT(t3,-1,lim);
	for(int i=0;i<n+m-1;i++)c[i]=(t3[i]%mod+mod)%mod; 
}
int d,n,m,k,A,B,C,f1[MAXN],f2[MAXN],f3[MAXN],dp[120005],tmp[120005],k1,k2; 
inline int calc2(int a,int b){
	//从 (0,0)-->(a,b) 无限制的方案数
	if((a+b)&1)return 0;
	return binom(a,(a+b)>>1); 
}

inline int calcf(int a,int b);
inline int calcg(int a,int b);

inline int calcf(int a,int b){//先接触 y=k1 
	if(a<abs(k1)+abs(b-k1))return 0;
	return dec(calc2(a,b),calcg(a,2*k2-b)); 
}
inline int calcg(int a,int b){//先接触 y=k2 
	if(a<abs(k2)+abs(b-k2))return 0;
	return dec(calc2(a,b),calcf(a,2*k1-b)); 
}
inline int solve(int a,int b,int K1,int K2){//(0,0)-->不接触 y=k1,y=k2; 
	k1=K1,k2=K2;
	return dec(calc2(a,b),add(calcf(a,2*k1-b),calcg(a,2*k2-b)));
}
inline void calc(int *Ans,int st,int ban){
	//只能在 [0,ban] 里面走
	//起点为 [0,st] 
	if(ban<=400){
		for(int i=0;i<=ban;i++)dp[i]=tmp[i]=0;
		dp[st]=1;		
		Ans[0]=1;
		for(int i=1;i<=d;i++){
			for(int j=0;j<=ban;j++){
				if(j-1>=0)tmp[j-1]=add(tmp[j-1],dp[j]);
				if(j+1<=ban)tmp[j+1]=add(tmp[j+1],dp[j]);
			}
			for(int j=0;j<=ban;j++){
				dp[j]=tmp[j],tmp[j]=0;
				Ans[i]=add(Ans[i],dp[j]);
			}
		}
	}else{
		Ans[0]=1;
		if(ban==0){
			for(int i=1;i<=d;i++)Ans[i]=1;
			return ;
		}
		for(int i=1;i<=d;i++){
			int res=Ans[i-1];
			//(0,st)->(i-1,0)
			//(0,st)->(i-1,ban)
			int s1=solve(i-1,-st,-1-st,ban+1-st),s2=solve(i-1,ban-st,-1-st,ban+1-st);
			res=dec(res,add(s1,s2));
			Ans[i]=add(1ll*res*2%mod,add(s1,s2)); 
			
		}
	}
}
signed main(){
	init_C(1000000);
	init();
	gi(d),gi(n),gi(m),gi(k),gi(A),gi(B),gi(C); 
	n--,m--,k--,A--,B--,C--;
	calc(f1,A,n);
	calc(f2,B,m);
	calc(f3,C,k);
	for(int i=0;i<=d;i++){
		f1[i]=1ll*f1[i]*inv[i]%mod;
		f2[i]=1ll*f2[i]*inv[i]%mod;
		f3[i]=1ll*f3[i]*inv[i]%mod;
	}
	int lim=init_NTT(2*d+2);
	MUL(f1,f2,f1,d+1,d+1,lim);
	MUL(f1,f3,f1,d+1,d+1,lim);
	pi(1ll*f1[d]*fac[d]%mod);
	return 0;
} 
/*
*/