AtCoder

f[i][j]表示向上扩展了i行，向右扩展了j列的方案数
显然f[i][j]=f[i-1][j]*j+f[i][j-1]*i;但是这样会有重复的
怎么去重是关键，top行和right列对称操作会重复算，所以最后为
f[i][j]=f[i-1][j]*j+f[i][j-1]*i-(i-1)*(j-1)*f[i-1][j-1];

#include <bits/stdc++.h>
#define inf 2333333333333333
#define N 3010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2020.7.21
using namespace std;
long long a,b,c,d,mod=998244353;
long long f[N][N];
void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

long long dfs(long long x,long long y){
    if(x<a || y<b) return 0;
    if(~f[x][y]) return f[x][y];
    long long ans=0;
    if(x==a) ans+=x*dfs(x,y-1),ans%=mod;
    if(y==b) ans+=y*dfs(x-1,y),ans%=mod;
    if(x>a && y>b) ans+=((x*dfs(x,y-1)%mod+y*dfs(x-1,y)%mod)%mod-(x-1)*(y-1)*dfs(x-1,y-1)%mod+mod)%mod,ans%=mod;
    f[x][y]=(ans+mod)%mod;
    return f[x][y];
}

signed main(){
    in(a);in(b);in(c);in(d);
    memset(f,-1,sizeof f);
    f[a][b]=1;
    o(dfs(c,d));
    return 0;
}